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<h2 class="hd hd-2 unit-title">1. Calculating derivatives</h2>
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<h3 class="hd hd-2">Calculating Derivatives: Finding formulas</h3>
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<h2 class="hd hd-2 unit-title">2. Objectives</h2>
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<p>
Slopes of tangent lines<br/>Limits<br/><span style="color:#27408C"><b class="bf">Formulas</b></span> </p><p><b class="bfseries">Objectives</b></p><p>
At the end of this sequence, and after some practice, you should be able to: </p><ul class="itemize"><li><p>
Find <span style="color:#99182C"><b class="bf">formulas for the derivatives</b></span> of simple functions. </p></li><li><p>
Know when to apply the <span style="color:#99182C"><b class="bf">power rule</b></span>. </p></li><li><p>
Apply <span style="color:#99182C"><b class="bf">linearity</b></span> to differential linear combinations. </p></li></ul><p><b class="bfseries">Contents: 18 pages</b></p><p>
13 videos (43 minutes 1x speed) 18 questions </p>
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<h2 class="hd hd-2 unit-title">3. Derivative of 2nd power</h2>
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<p>
In the last sequence, you learned how to sketch the graph of [mathjaxinline]\ f'[/mathjaxinline] given the graph of [mathjaxinline]\ f[/mathjaxinline]. Now we would like to find formulas. We'll start by computing the derivative of [mathjaxinline]\ f(x) = x^2[/mathjaxinline]. Before we do this, let's try to anticipate what [mathjaxinline]\ f'[/mathjaxinline] should look like. </p>
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Sketch graph of derivative
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Here's the graph of [mathjaxinline]\ f(x) = x^2[/mathjaxinline]. </p>
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Use the tools below to sketch what you suspect the graph of [mathjaxinline]\ f'[/mathjaxinline] looks like. </p>
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(This graph need only be qualitatively correct &#8211; the actual values are not important here. Where does the graph of the derivative cross the [mathjaxinline]x[/mathjaxinline]-axis?) </p>
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Definition of derivative at a point
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Now that we have a sense of what we should get, we'll compute it exactly. Let's remember the definition of the derivative at a point [mathjaxinline]\ f'(a)[/mathjaxinline]: </p>
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The definition of the derivative at the point [mathjaxinline]x=3[/mathjaxinline] is: </p>
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<td class="equation" style="width:80%; border:none">[mathjax]f'(3) = \lim _{b\rightarrow \underline{\; \; \; }} \frac{f(b)-f(3)}{\underline{\; \; \; \; \; \; }}.[/mathjax]</td>
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<h2 class="hd hd-2 unit-title">4. Derivatives of linear functions</h2>
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<h3 class="hd hd-2">Derivatives of Linear Functions</h3>
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Limit of Delta x
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As [mathjaxinline]b[/mathjaxinline] gets closer and closer to [mathjaxinline]x[/mathjaxinline], what is [mathjaxinline]\Delta x[/mathjaxinline] getting closer and closer to? </p>
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<h3 class="hd hd-2">Using the Delta definition</h3>
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Derivative definition practice
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Let [mathjaxinline]h(x) = 1/x^2[/mathjaxinline]. Using whichever notation for the definition of the derivative you are <span style="color:#99182C"><b class="bf">less comfortable</b></span> with, calculate [mathjaxinline]h'(x)[/mathjaxinline]. </p>
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(Type [mathjaxinline]*[/mathjaxinline] for multiplication; e.g. 2[mathjaxinline]*[/mathjaxinline]x for [mathjaxinline]2x[/mathjaxinline]. Type / for division; e.g. 2/x for [mathjaxinline]\displaystyle \frac{2}{x}[/mathjaxinline]. Type [mathjaxinline]\wedge[/mathjaxinline] for exponents; e.g. x[mathjaxinline]\wedge[/mathjaxinline]2 for [mathjaxinline]x^2[/mathjaxinline].) </p>
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<h2 class="hd hd-2 unit-title">6. Linearity of the derivative</h2>
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<p>
So far we've calculated derivatives of the following functions using the limit definition of the derivative. </p><table id="a0000000072" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000073"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle f(x) = x^2[/mathjaxinline]
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[mathjaxinline]\displaystyle \rightarrow[/mathjaxinline]
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[mathjaxinline]\displaystyle f'(x) = 2x[/mathjaxinline]
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[mathjaxinline]\displaystyle g(x) = mx + b[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightarrow[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle g'(x) = m[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000075"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle h(x) = \frac{1}{x}[/mathjaxinline]
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[mathjaxinline]\displaystyle \rightarrow[/mathjaxinline]
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[mathjaxinline]\displaystyle h'(x) = -\frac{1}{x^2}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Now we are going to use these formulas to find derivatives of lots of related functions. One nice thing is that we won't have to use the limit definition for these functions! </p>
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Relationship between graphs
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What do we mean by &#8220;related"? Well, let's take an example. Suppose we know the derivative of [mathjaxinline]\ f(x)[/mathjaxinline], and then we have a new function [mathjaxinline]g[/mathjaxinline], where [mathjaxinline]g(x) = 2f(x)[/mathjaxinline] for all inputs [mathjaxinline]x[/mathjaxinline]. </p>
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What do we do to the graph of [mathjaxinline]\ f[/mathjaxinline] in order to get the graph of [mathjaxinline]g[/mathjaxinline]? </p>
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Relationship between derivatives
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If [mathjaxinline]g(x) = 2f(x)[/mathjaxinline], the graph of [mathjaxinline]g[/mathjaxinline] is the graph of [mathjaxinline]\ f[/mathjaxinline] stretched vertically by a factor of 2. So what is the relationship between the slope of the tangent line to the graph of [mathjaxinline]g[/mathjaxinline] at a point [mathjaxinline]x=a[/mathjaxinline] and the slope of the tangent line to the graph of [mathjaxinline]\ f[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline]? </p>
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<h3 class="hd hd-2">Derivatives of constant multiples</h3>
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Differentiation practice
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Let [mathjaxinline]\ f(x) = \displaystyle {-\frac{3}{x}}.[/mathjaxinline] Find [mathjaxinline]\ f'(x)[/mathjaxinline]. </p>
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(Type [mathjaxinline]*[/mathjaxinline] for multiplication; e.g. 2[mathjaxinline]*[/mathjaxinline]x for [mathjaxinline]2x[/mathjaxinline]. Type / for division; e.g. 2/x for [mathjaxinline]\displaystyle \frac{2}{x}[/mathjaxinline]. Type [mathjaxinline]\wedge[/mathjaxinline] for exponents; e.g. x[mathjaxinline]\wedge[/mathjaxinline]2 for [mathjaxinline]x^2[/mathjaxinline].) </p>
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<p style="display:inline">[mathjaxinline]\ f'(x)=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">8. Derivatives of constant multiples proof</h2>
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Suppose that [mathjaxinline]g(x) = kf(x)[/mathjaxinline] for all [mathjaxinline]x[/mathjaxinline], where [mathjaxinline]k[/mathjaxinline] is a constant. We want to prove that [mathjaxinline]g'(x) = kf'(x)[/mathjaxinline] at any point [mathjaxinline]x[/mathjaxinline] where [mathjaxinline]\ f[/mathjaxinline] is differentiable. </p><p>
We know that </p><table id="a0000000077" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000078"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle g'(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} \frac{g(x+ \Delta x) - g(x)}{\Delta x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000079"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} \frac{kf(x + \Delta x) - kf(x)}{\Delta x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000080"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} k\frac{f(x + \Delta x) - f(x)}{\Delta x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000081"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} k \ \lim _{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
The first limit is just [mathjaxinline]k[/mathjaxinline], and the second limit is the definition of [mathjaxinline]\ f'(x)[/mathjaxinline]. So we get [mathjaxinline]g'(x) = k f'(x)[/mathjaxinline]. </p>
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More differentiation practice
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<p>
Let [mathjaxinline]\ f(x) = x^2 - \displaystyle {\frac{1}{x}}.[/mathjaxinline] Find [mathjaxinline]\ f'(x)[/mathjaxinline]. </p>
<p>
(Type [mathjaxinline]*[/mathjaxinline] for multiplication; e.g. 2[mathjaxinline]*[/mathjaxinline]x for [mathjaxinline]2x[/mathjaxinline]. Type / for division; e.g. 2/x for [mathjaxinline]\displaystyle \frac{2}{x}[/mathjaxinline]. Type [mathjaxinline]\wedge[/mathjaxinline] for exponents; e.g. x[mathjaxinline]\wedge[/mathjaxinline]2 for [mathjaxinline]x^2[/mathjaxinline].) </p>
<p>
<p style="display:inline">[mathjaxinline]\ f'(x)=[/mathjaxinline]</p>
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In the last video we told you that the derivative of a sum was the sum of derivatives, but we didn't justify it. We'll do that in this next video. </p>
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<h3 class="hd hd-2">Derivative of a sum - proof</h3>
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<h2 class="hd hd-2 unit-title">10. Derivatives and linearity</h2>
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If [mathjaxinline]g(x) = k\ f(x)[/mathjaxinline] for some constant [mathjaxinline]k[/mathjaxinline], then </p><table id="a0000000083" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle g'(x) = k\ f'(x)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
at all points where [mathjaxinline]\ f[/mathjaxinline] is differentiable. </p><p>
If [mathjaxinline]h(x) = f(x) + g(x)[/mathjaxinline], then </p><table id="a0000000084" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle h'(x) = f'(x) + g'(x)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
at all points where [mathjaxinline]\ f[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline] are differentiable. </p><p>
Similarly, if [mathjaxinline]j(x) = f(x) - g(x)[/mathjaxinline], then </p><table id="a0000000085" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle j'(x) = f'(x) - g'(x)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
at all points where [mathjaxinline]\ f[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline] are differentiable. </p>
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<h2 class="hd hd-2 unit-title">11. Linearity examples</h2>
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We've seen that differentiation “respects" addition and multiplication by a constant. That is, if you take a derivative of a sum of functions, you get the same thing as if you differentiated each part, and then added the derivatives. Similarly, if you take the derivative of [mathjaxinline]k[/mathjaxinline] times a function, where [mathjaxinline]k[/mathjaxinline] is a constant, then you get [mathjaxinline]k[/mathjaxinline] times the derivative of the original function. </p><p>
Respecting addition and constant multiplication in this way is called “linearity," and it is an important property of the derivative operation! We combine both parts of linearity in the following problems. </p>
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Pumpkin velocity
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<p>
Remember our pumpkin that we threw from a building. </p>
<p>
The height was [mathjaxinline]\ f(t) = 100+20t-5t^2[/mathjaxinline] meters above the ground after [mathjaxinline]t[/mathjaxinline] seconds. We can now calculate the velocity [mathjaxinline]\ f'[/mathjaxinline] at any time [mathjaxinline]t[/mathjaxinline]. What is [mathjaxinline]\ f'(t)[/mathjaxinline]? </p>
<p>
(Type [mathjaxinline]*[/mathjaxinline] for multiplication; e.g. 2[mathjaxinline]*[/mathjaxinline]t for [mathjaxinline]2t[/mathjaxinline]. Type / for division; e.g. 2/t for [mathjaxinline]\displaystyle \frac{2}{t}[/mathjaxinline]. Type [mathjaxinline]\wedge[/mathjaxinline] for exponents; e.g. t[mathjaxinline]\wedge[/mathjaxinline]2 for [mathjaxinline]t^2[/mathjaxinline].) </p>
<p>
<p style="display:inline">[mathjaxinline]\ f'(t)=[/mathjaxinline]</p>
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Rate of price increase
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In this problem, we want to compute the rate of change of the price of a pizza party! </p>
<p>
Suppose that the price of a pizza in 2014 is $12, and the rate of change of the price of a pizza is [mathjaxinline]0.2[/mathjaxinline] dollars per year. </p>
<p>
Meanwhile, a bottle of Coca-Cola costs $2 in 2014, and its price is rising at a rate of [mathjaxinline]0.1[/mathjaxinline] dollars per year. </p>
<p>
Since the prices of pizza and Coke change with time, so does the cost of a party. Let [mathjaxinline]h(t)[/mathjaxinline] be the cost of 10 pizzas and 5 bottles of Coca-Cola in year [mathjaxinline]t[/mathjaxinline]. That is, [mathjaxinline]h(t) = 10f(t) + 5g(t)[/mathjaxinline], where [mathjaxinline]\ f(t)[/mathjaxinline] is the price of a pizza in year [mathjaxinline]t[/mathjaxinline], and [mathjaxinline]g(t)[/mathjaxinline] is the price of a bottle of Coke in year [mathjaxinline]t[/mathjaxinline]. </p>
<p>
How fast is the cost of the pizza party growing? In other words, what is [mathjaxinline]h'(2014)[/mathjaxinline]? </p>
<p>
<p style="display:inline">[mathjaxinline]h'(2014)=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">12. Linearity example</h2>
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Suppose [mathjaxinline]\ f(x) = -3x^2 + \frac{1}{x} - 2[/mathjaxinline] and we want to calculate [mathjaxinline]\ f'(1)[/mathjaxinline]. </p>
<p>
First, let's think about how one might go about solving such a problem. One idea would be to calculate [mathjaxinline]\ f'(x)[/mathjaxinline] using the sum and constant multiple rules we've developed, and then plug in [mathjaxinline]x=1[/mathjaxinline] to find [mathjaxinline]\ f'(1)[/mathjaxinline]. Another idea would be to calculate [mathjaxinline]\ f(1)[/mathjaxinline] first, and then take the derivative of the result. Which method is correct? </p>
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<text> Calculate [mathjaxinline]\ f'(x)[/mathjaxinline] first, then plug in [mathjaxinline]x=1[/mathjaxinline] to find [mathjaxinline]\ f'(1)[/mathjaxinline]</text>
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<text> Plug in [mathjaxinline]x=1[/mathjaxinline] to find [mathjaxinline]\ f(1)[/mathjaxinline] first, then take the derivative of the result</text>
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<h2 class="hd hd-2 unit-title">13. Power Rule videos</h2>
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<h3 class="hd hd-2">Power Rule</h3>
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<a class="btn btn-link" href="/courses/course-v1:MITx+18.01.1x+2T2019/xblock/block-v1:MITx+18.01.1x+2T2019+type@video+block@der_4-tab13-video1/handler/transcript/download" data-value="srt">Download SubRip (.srt) file</a>
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<a class="btn btn-link" href="/courses/course-v1:MITx+18.01.1x+2T2019/xblock/block-v1:MITx+18.01.1x+2T2019+type@video+block@der_4-tab13-video1/handler/transcript/download" data-value="txt">Download Text (.txt) file</a>
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<div class="xblock xblock-public_view xblock-public_view-video xmodule_display xmodule_VideoBlock" data-request-token="fb3593a4ea7e11efb8120e0e3c45b88f" data-usage-id="block-v1:MITx+18.01.1x+2T2019+type@video+block@der_4-tab13-video2" data-graded="True" data-has-score="False" data-runtime-version="1" data-block-type="video" data-init="XBlockToXModuleShim" data-course-id="course-v1:MITx+18.01.1x+2T2019" data-runtime-class="LmsRuntime">
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<h3 class="hd hd-2">Factoring the difference of nth powers (optional)</h3>
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<a class="btn btn-link" href="/courses/course-v1:MITx+18.01.1x+2T2019/xblock/block-v1:MITx+18.01.1x+2T2019+type@video+block@der_4-tab13-video2/handler/transcript/download" data-value="txt">Download Text (.txt) file</a>
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<h2 class="hd hd-2 unit-title">14. Power Rule examples</h2>
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Power Rule example 1
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What is the derivative of [mathjaxinline]\ f(x) = x^{10}[/mathjaxinline]? </p>
<p>
(Type [mathjaxinline]*[/mathjaxinline] for multiplication; e.g. 2[mathjaxinline]*[/mathjaxinline]x for [mathjaxinline]2x[/mathjaxinline]. Type [mathjaxinline]\wedge[/mathjaxinline] for exponents; e.g. x[mathjaxinline]\wedge[/mathjaxinline]2 for [mathjaxinline]x^2[/mathjaxinline].) </p>
<p>
<p style="display:inline">[mathjaxinline]\ f'(x)=[/mathjaxinline]</p>
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Power Rule example 2
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What is the derivative of [mathjaxinline]g(x) = 5x^{6} - \frac{x^4}{2} - 7[/mathjaxinline]? </p>
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(Type [mathjaxinline]*[/mathjaxinline] for multiplication; e.g. 2[mathjaxinline]*[/mathjaxinline]x for [mathjaxinline]2x[/mathjaxinline]. Type [mathjaxinline]\wedge[/mathjaxinline] for exponents; e.g. x[mathjaxinline]\wedge[/mathjaxinline]2 for [mathjaxinline]x^2[/mathjaxinline].) </p>
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<p style="display:inline">[mathjaxinline]g'(x)=[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_der_4-tab14-problem2_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<h2 class="hd hd-2 unit-title">15. Extended Power Rule</h2>
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<h3 class="hd hd-2">Extended Power Rule</h3>
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<h2 class="hd hd-2 unit-title">16. General Power Rule questions</h2>
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Power Rule limitations
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Which of the following functions does the power rule directly apply to? </p>
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<text>[mathjaxinline]p(x) = (1+2x)^{10}[/mathjaxinline]</text>
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<text>[mathjaxinline]q(x) = u^5[/mathjaxinline], where [mathjaxinline]u = \sin x[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">17. Recitation video</h2>
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<p><b class="bfseries"><span style="color:#FF7F00">Video note:</span></b> At 2:23 the Joel accidentally says that the derivative is [mathjaxinline]3x^2+1[/mathjaxinline] but means [mathjaxinline]3x^2-1[/mathjaxinline]. </p>
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<h2 class="hd hd-2 unit-title">18. Distinguishing between tangent line and derivative</h2>
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Suppose that the tangent line to the graph of [mathjaxinline]\ f[/mathjaxinline] at [mathjaxinline]x=5[/mathjaxinline] has equation [mathjaxinline]y = 4x + 3[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\ f(5)=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">19. Summary</h2>
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<p><b class="bfseries">Derivatives of constant multiples</b></p><p>
If [mathjaxinline]g(x) = k\ f(x)[/mathjaxinline] for some constant [mathjaxinline]k[/mathjaxinline], then </p><table id="a0000000088" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle g'(x) = k\ f'(x)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
at all points where [mathjaxinline]\ f[/mathjaxinline] is differentiable. </p><p><b class="bfseries">Derivatives of sums</b></p><p>
If [mathjaxinline]h(x) = f(x) + g(x)[/mathjaxinline], then </p><table id="a0000000089" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle h'(x) = f'(x) + g'(x)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
at all points where [mathjaxinline]\ f[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline] are differentiable. </p><p><b class="bfseries">Derivatives of differences</b></p><p>
Similarly, if [mathjaxinline]j(x) = f(x) - g(x)[/mathjaxinline], then </p><table id="a0000000090" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle j'(x) = f'(x) - g'(x)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
at all points where [mathjaxinline]\ f[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline] are differentiable. </p><p><b class="bfseries">Derivatives of constant multiples - proof</b></p><p>
Suppose that [mathjaxinline]g(x) = kf(x)[/mathjaxinline] for all [mathjaxinline]x[/mathjaxinline], where [mathjaxinline]k[/mathjaxinline] is a constant. We want to prove that [mathjaxinline]g'(x) = kf'(x)[/mathjaxinline] at any point [mathjaxinline]x[/mathjaxinline] where [mathjaxinline]\ f[/mathjaxinline] is differentiable. </p><p>
We know that </p><table id="a0000000091" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000092"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle g'(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} \frac{g(x+ \Delta x) - g(x)}{\Delta x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000093"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} \frac{kf(x + \Delta x) - kf(x)}{\Delta x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000094"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} k\frac{f(x + \Delta x) - f(x)}{\Delta x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000095"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = \lim _{\Delta x \rightarrow 0} k \ \lim _{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
The first limit is just [mathjaxinline]k[/mathjaxinline], and the second limit is the definition of [mathjaxinline]\ f'(x)[/mathjaxinline]. So we get [mathjaxinline]g'(x) = k f'(x)[/mathjaxinline]. </p><p><b class="bfseries">What is linearity?</b></p><p>
We've seen that differentiation “respects" addition and multiplication by a constant. That is, if you take a derivative of a sum of functions, you get the same thing as if you differentiated each part, and then added the derivatives. Similarly, if you take the derivative of [mathjaxinline]k[/mathjaxinline] times a function, where [mathjaxinline]k[/mathjaxinline] is a constant, then you get [mathjaxinline]k[/mathjaxinline] times the derivative of the original function. </p><p>
Respecting addition and constant multiplication in this way is called “linearity," and it is an important property of the derivative operation! </p><p><b class="bfseries">The Power Rule</b></p><p>
If [mathjaxinline]n[/mathjaxinline] is any fixed number, and [mathjaxinline]\ f(x) = x^ n[/mathjaxinline], then [mathjaxinline]\ f'(x) = n x^{n-1}[/mathjaxinline]. </p>
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