009A Sample Midterm 3, Problem 3 Detailed Solution

Let  $y={\sqrt {2x+5}},x\geq 0.$

(a) Use the definition of the derivative to compute   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {dy}{dx}}.}

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

---

Background Information:
Recall
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$

Solution:

(a)

Step 1:
Let  $f(x)=3{\sqrt {2x+5}}.$
Using the limit definition of the derivative, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3{\sqrt {2(x+h)+5}}-3{\sqrt {2x+5}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3{\sqrt {2x+2h+5}}-3{\sqrt {2x+5}}}{h}}}\\&&\\&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {{\sqrt {2x+2h+5}}-{\sqrt {2x+5}}}{h}}.}\end{array}}$

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {({\sqrt {2x+2h+5}}-{\sqrt {2x+5}})}{h}}{\frac {({\sqrt {2x+2h+5}}+{\sqrt {2x+5}})}{({\sqrt {2x+2h+5}}+{\sqrt {2x+5}})}}}\\&&\\&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {(2x+2h+5)-(2x+5)}{h({\sqrt {2x+2h+5}}+{\sqrt {2x+5}})}}}\\&&\\&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {2h}{h({\sqrt {2x+2h+5}}+{\sqrt {2x+5}})}}}\\&&\\&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {2}{{\sqrt {2x+2h+5}}+{\sqrt {2x+5}}}}}\\&&\\&=&\displaystyle {3{\frac {2}{{\sqrt {2x+5}}+{\sqrt {2x+5}}}}}\\&&\\&=&\displaystyle {{\frac {3}{\sqrt {2x+5}}}.}\end{array}}$

(b)

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

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 \frac{3}{\sqrt{2x+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)=\cos(\cos(e^x))(-\sin(e^x))(e^x)}

Return to Sample Exam