009A Sample Midterm 2, Problem 5

Find the derivatives of the following functions. Do not simplify.

(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 f(x)=\tan^3(7x^2+5) }

(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 g(x)=\sin(\cos(e^x)) }

(c)   $h(x)={\frac {(5x^{2}+7x)^{3}}{\ln(x^{2}+1)}}$

Foundations:
1. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$
2. Trig Derivatives
${\frac {d}{dx}}(\sin x)=\cos x,\quad {\frac {d}{dx}}(\cos x)=-\sin x$
3. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$
4. Derivative of natural logarithm
${\frac {d}{dx}}(\ln x)={\frac {1}{x}}$

Solution:

(a)

Step 1:
First, we use the Chain Rule to get
$f'(x)=3\tan ^{2}(7x^{2}+5)(\tan(7x^{2}+5))'.$

Step 2:
Now, we use the Chain Rule again to get
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {3\tan ^{2}(7x^{2}+5)(\tan(7x^{2}+5))'}\\&&\\&=&\displaystyle {3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(7x^{2}+5)'}\\&&\\&=&\displaystyle {3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(14x).}\end{array}}$

(b)

Step 1:
First, we use the Chain Rule to get
$g'(x)=\cos(\cos(e^{x}))(\cos(e^{x}))'.$

Step 2:
Now, we use the Chain Rule again to get
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\cos(\cos(e^{x}))(\cos(e^{x}))'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x})'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x}).}\end{array}}$

(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}}

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