Difference between revisions of "Chain Rule"
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!Solution: | !Solution: | ||
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| − | |Let <math>f(x)=\sin (x)</math> and <math>g(x)=3x.</math> | + | |Let <math style="vertical-align: -5px">f(x)=\sin (x)</math> and <math style="vertical-align: -5px">g(x)=3x.</math> |
|- | |- | ||
| − | |Then, <math>f'(x)=\cos(x)</math> and <math>g'(x)=3.</math> | + | |Then, <math style="vertical-align: -5px">f'(x)=\cos(x)</math> and <math style="vertical-align: -5px">g'(x)=3.</math> |
|- | |- | ||
| − | |Now, <math>h(x)=f(g(x)).</math> | + | |Now, <math style="vertical-align: -6px">h(x)=f(g(x)).</math> |
|- | |- | ||
|Using the Chain Rule, we have | |Using the Chain Rule, we have | ||
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!Solution: | !Solution: | ||
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| − | |Let <math>f(x)=x^8</math> and <math>g(x)=x+1.</math> | + | |Let <math style="vertical-align: -5px">f(x)=x^8</math> and <math style="vertical-align: -5px">g(x)=x+1.</math> |
|- | |- | ||
| − | |Then, <math>f'(x)=8x^7</math> and <math>g'(x)=1.</math> | + | |Then, <math style="vertical-align: -5px">f'(x)=8x^7</math> and <math style="vertical-align: -5px">g'(x)=1.</math> |
|- | |- | ||
| − | |Now, <math>h(x)=f(g(x)).</math> | + | |Now, <math style="vertical-align: -6px">h(x)=f(g(x)).</math> |
|- | |- | ||
|Using the Chain Rule, we have | |Using the Chain Rule, we have | ||
| Line 90: | Line 90: | ||
!Solution: | !Solution: | ||
|- | |- | ||
| − | |Let <math>f(x)=\ln (x)</math> and <math>g(x)=x^2.</math> | + | |Let <math style="vertical-align: -5px">f(x)=\ln (x)</math> and <math style="vertical-align: -5px">g(x)=x^2.</math> |
|- | |- | ||
| − | |Then, <math>f'(x)=\frac{1}{x}</math> and <math>g'(x)=2x.</math> | + | |Then, <math style="vertical-align: -13px">f'(x)=\frac{1}{x}</math> and <math style="vertical-align: -5px">g'(x)=2x.</math> |
|- | |- | ||
| − | |Now, <math>h(x)=f(g(x)).</math> | + | |Now, <math style="vertical-align: -6px">h(x)=f(g(x)).</math> |
|- | |- | ||
|Using the Chain Rule, we have | |Using the Chain Rule, we have | ||
Revision as of 09:02, 9 October 2017
Introduction
It is relatively easy to calculate the derivatives of simple functions, like polynomials or trigonometric functions.
But, what about more complicated functions?
For example, or
Well, the key to calculating the derivatives of these functions is to recognize that these functions are compositions.
For it is the composition of the function with
Similarly, for it is the composition of and
So, how do we take the derivative of compositions?
The answer to this question is exactly the Chain Rule.
Chain Rule
Let be a differentiable function of and let be a differentiable function of
Then, is a differentiable function of and
Warm-Up
Calculate
1)
| Solution: |
|---|
| Let and |
| Then, and |
| Now, |
| Using the Chain Rule, we have |
|
| Final Answer: |
|---|
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=3\cos (3x)} |
2) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=(x+1)^8}
| Solution: |
|---|
| Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x^8} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=x+1.} |
| Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=8x^7} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=1.} |
| Now, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=f(g(x)).} |
| Using the Chain Rule, we have |
|
| Final Answer: |
|---|
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=8(x+1)^7.} |
3) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\ln(x^2)}
| Solution: |
|---|
| Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\ln (x)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=x^2.} |
| Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{1}{x}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=2x.} |
| Now, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=f(g(x)).} |
| Using the Chain Rule, we have |
|
| Final Answer: |
|---|
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{2}{x}} |
Exercise 1
Calculate the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\frac{1}{x^2}(\csc x-4).}
First, we need to know the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \csc x.} Recall
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \csc x =\frac{1}{\sin x}.}
Now, using the Quotient Rule, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\ &&\\ & = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\ &&\\ & = & \displaystyle{-\csc x \cot x.} \end{array}}
Using the Product Rule and Power Rule, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.} \end{array}}
So, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
Exercise 2
Calculate the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=2x\sin x \sec x.}
Notice that the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)} is the product of three functions.
We start by grouping two of the functions together. So, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=(2x\sin x)\sec x.}
Using the Product Rule, we get
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\ &&\\ & = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.} \end{array}}
Now, we need to use the Product Rule again. So,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\ &&\\ & = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.} \end{array}}
So, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
But, there is another way to do this problem. Notice
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\ &&\\ & = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\ &&\\ & = & \displaystyle{2x\tan x.} \end{array}}
Now, you would only need to use the Product Rule once instead of twice.
Exercise 3
Calculate the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.}
Using the Quotient Rule, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.}
Now, we need to use the Product Rule. So, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\ &&\\ & = & \displaystyle{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.} \end{array}}
So, we get
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}
Exercise 4
Calculate the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\frac{e^x}{x^2\sin x}.}
First, using the Quotient Rule, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.} \end{array}}
Now, we need to use the Product Rule. So, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.} \end{array}}
So, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}