Difference between revisions of "009A Sample Midterm 2, Problem 5"
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<span class="exam">Find the derivatives of the following functions. Do not simplify. | <span class="exam">Find the derivatives of the following functions. Do not simplify. | ||
| − | <span class="exam">(a) <math>f(x)=\tan^3(7x^2+5) </math> | + | <span class="exam">(a) <math style="vertical-align: -5px">f(x)=\tan^3(7x^2+5) </math> |
| − | <span class="exam">(b) <math>g(x)=\sin(\cos(e^x)) </math> | + | <span class="exam">(b) <math style="vertical-align: -5px">g(x)=\sin(\cos(e^x)) </math> |
| − | <span class="exam">(c) <math>h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math> | + | <span class="exam">(c) <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math> |
Revision as of 16:21, 18 February 2017
Find the derivatives of the following functions. Do not simplify.
(a)
(b)
(c) 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{(5x^2+7x)^3}{\ln(x^2+1)} }
| Foundations: |
|---|
| 1. Chain Rule |
| 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 \frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)} |
| 2. Trig Derivatives |
| 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 \frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\cos x)=-\sin x} |
| 3. Quotient Rule |
| 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 \frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}} |
| 4. Derivative of natural logarithm |
| 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 \frac{d}{dx}(\ln x)=\frac{1}{x}} |
Solution:
(a)
| Step 1: |
|---|
| First, we use the Chain Rule to 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 f'(x)=3\tan^2(7x^2+5)(\tan(7x^2+5))'.} |
| Step 2: |
|---|
| Now, we use the Chain Rule again to 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{f'(x)} & = & \displaystyle{3\tan^2(7x^2+5)(\tan(7x^2+5))'}\\ &&\\ & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(7x^2+5)'}\\ &&\\ & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x).} \end{array}} |
(b)
| Step 1: |
|---|
| First, we use the Chain Rule to 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 g'(x)=\cos(\cos(e^x))(\cos(e^x))'.} |
| Step 2: |
|---|
| Now, we use the Chain Rule again to 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{\cos(\cos(e^x))(\cos(e^x))'}\\ &&\\ & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x)'}\\ &&\\ & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x).} \end{array}} |
(c)
| Step 1: |
|---|
| First, we use the Quotient Rule to 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{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}.} |
| Step 2: |
|---|
| Now, we use the Chain Rule to 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{h'(x)} & = & \displaystyle{\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}}\\ &&\\ & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(5x^2+7x)'-(5x^2+7x)^2\frac{1}{x^2+1}(x^2+1)'}{(\ln(x^2+1))^2}}\\ &&\\ & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}.} \end{array}} |
| Final Answer: |
|---|
| (a) 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 3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)} |
| (b) 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 \cos(\cos(e^x))(-\sin(e^x))(e^x)} |
| (c) 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 \frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}} |