Difference between revisions of "Chain Rule"
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 32: | Line 32: | ||
!Solution: | !Solution: | ||
|- | |- | ||
| − | | | + | |Let <math>f(x)=\sin (x)</math> and <math>g(x)=3x.</math> |
|- | |- | ||
| − | | | + | |Then, <math>f'(x)=\cos(x)</math> and <math>g'(x)=3.</math> |
| − | |||
|- | |- | ||
| − | | | + | |Now, <math>h(x)=f(g(x)).</math> |
|- | |- | ||
| − | | | + | |Using the Chain Rule, we have |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| | | | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | \displaystyle{ | + | \displaystyle{h'(x)} & = & \displaystyle{f'(g(x))\cdot g'(x)}\\ |
| − | |||
| − | |||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{ | + | & = & \displaystyle{\cos (3x)\cdot 3}\\ |
&&\\ | &&\\ | ||
| − | & = & \displaystyle{ | + | & = & \displaystyle{3\cos (3x).} |
\end{array}</math> | \end{array}</math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 71: | Line 53: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math> | + | | <math>h'(x)=3\cos (3x)</math> |
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 83: | Line 61: | ||
!Solution: | !Solution: | ||
|- | |- | ||
| − | | | + | |Let <math>f(x)=x^8</math> and <math>g(x)=x+1.</math> |
| − | |||
|- | |- | ||
| − | | | + | |Then, <math>f'(x)=8x^7</math> and <math>g'(x)=1.</math> |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| − | |< | + | |Now, <math>h(x)=f(g(x)).</math> |
|- | |- | ||
| − | | | + | |Using the Chain Rule, we have |
| − | |||
| − | |||
|- | |- | ||
| | | | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | \displaystyle{ | + | \displaystyle{h'(x)} & = & \displaystyle{f'(g(x))\cdot g'(x)}\\ |
&&\\ | &&\\ | ||
| − | & = & \displaystyle{ | + | & = & \displaystyle{8(x+1)^7\cdot 1}\\ |
&&\\ | &&\\ | ||
| − | & = & \displaystyle{x+ | + | & = & \displaystyle{8(x+1)^7.} |
\end{array}</math> | \end{array}</math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 118: | Line 82: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | || <math> | + | || <math>h'(x)=8(x+1)^7.</math> |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 132: | Line 90: | ||
!Solution: | !Solution: | ||
|- | |- | ||
| − | |Using the | + | |Let <math>f(x)=\ln (x)</math> and <math>g(x)=x^2.</math> |
| + | |- | ||
| + | |Then, <math>f'(x)=\frac{1}{x}</math> and <math>g'(x)=2x.</math> | ||
| + | |- | ||
| + | |Now, <math>h(x)=f(g(x)).</math> | ||
| + | |- | ||
| + | |Using the Chain Rule, we have | ||
|- | |- | ||
| | | | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | \displaystyle{ | + | \displaystyle{h'(x)} & = & \displaystyle{f'(g(x))\cdot g'(x)}\\ |
| − | |||
| − | |||
| − | |||
| − | |||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{\frac{1}{ | + | & = & \displaystyle{\frac{1}{x^2}\cdot 2x}\\ |
&&\\ | &&\\ | ||
| − | & = & \displaystyle{\ | + | & = & \displaystyle{\frac{2}{x}.} |
\end{array}</math> | \end{array}</math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 158: | Line 111: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math> | + | | <math>h'(x)=\frac{2}{x}</math> |
|- | |- | ||
|} | |} | ||
Revision as of 08:47, 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, 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)=\sin(3x)} or 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)^8?}
Well, the key to calculating the derivatives of these functions is to recognize that these functions are compositions.
For 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)=\sin(3x),} it is the composition of 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 y=3x} with 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 y=\sin(x).}
Similarly, for 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)^8,} it is the composition 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 y=x+1} 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 y=x^8.}
So, how do we take the derivative of compositions?
The answer to this question is exactly the Chain Rule.
Chain Rule
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 y=f(u)} be a differentiable function 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 u} and 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 u=g(x)} be a differentiable function 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 x.}
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 y=f(g(x))} is a differentiable function 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 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 y'=f'(g(x))\cdot g'(x).}
Warm-Up
Calculate 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).}
1) 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)=\sin(3x)}
| 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)=\sin (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)=3x.} |
| 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)=\cos(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)=3.} |
| 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)=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,
So, we have
But, there is another way to do this problem. Notice
Now, you would only need to use the Product Rule once instead of twice.
Exercise 3
Calculate the derivative of
Using the Quotient Rule, we have
Now, we need to use the Product Rule. So, we have
So, we get
Exercise 4
Calculate the derivative of
First, using the Quotient Rule, we have
Now, we need to use the Product Rule. So, we have
So, we have