009A Sample Midterm 3, Problem 3 Detailed 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 y=\sqrt{2x+5},x\ge 0.}

(a) Use the definition of the derivative to compute   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{dy}{dx}.}

(b) Find the equation of the tangent line to  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=3\sqrt{2x+5}}   at  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 (2,9).}

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Background Information:
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 f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}

Solution:

(a)

Step 1:
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)=3\sqrt{-2x+5}.}
Using the limit definition of the derivative, 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{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2(x+h)+5}-3\sqrt{-2x+5}}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2x+-2h+5}-3\sqrt{-2x+5}}{h}}\\ &&\\ & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{\sqrt{-2x+-2h+5}-\sqrt{-2x+5}}{h}.} \end{array}}

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, 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{3\lim_{h\rightarrow 0} \frac{(\sqrt{-2x+-2h+5}-\sqrt{-2x+5})}{h} \frac{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\ &&\\ & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(-2x+-2h+5)-(-2x+5)}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\ &&\\ & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2h}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\ &&\\ & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2}{\sqrt{-2x+-2h+5}+\sqrt{-2x+5}}}\\ &&\\ & = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\ &&\\ & = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.} \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}}

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 f'(x)=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 g'(x)=\cos(\cos(e^x))(-\sin(e^x))(e^x)}

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