# 009A Sample Midterm 3, Problem 3 Detailed Solution

Let  ${\displaystyle y=3{\sqrt {2x+5}},x\geq 0.}$

(a) Use the definition of the derivative to compute   ${\displaystyle {\frac {dy}{dx}}.}$

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

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

Solution:

(a)

Step 1:
Let  ${\displaystyle f(x)=3{\sqrt {2x+5}}.}$
Using the limit definition of the derivative, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3{\sqrt {2(x+h)+5}}-3{\sqrt {2x+5}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3{\sqrt {2x+2h+5}}-3{\sqrt {2x+5}}}{h}}}\\&&\\&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {{\sqrt {2x+2h+5}}-{\sqrt {2x+5}}}{h}}.}\end{array}}}$

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

(b)

Step 1:
We start by finding the slope of the tangent line to  ${\displaystyle f(x)=3{\sqrt {2x+5}}}$  at  ${\displaystyle (2,9).}$
Using the derivative calculated in part (a), the slope is
${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {\frac {3}{\sqrt {2(2)+5}}}\\&&\\&=&\displaystyle {\frac {3}{\sqrt {9}}}\\&&\\&=&\displaystyle {1.}\end{array}}}$
Step 2:
Now, the tangent line to  ${\displaystyle f(x)=3{\sqrt {2x+5}}}$  at  ${\displaystyle (2,9)}$
has slope  ${\displaystyle m=1}$  and passes through the point  ${\displaystyle (2,9).}$
Hence, the equation of this line is
${\displaystyle y=(x-2)+9.}$
If we simplify, we get
${\displaystyle y=x+7.}$

(a)     ${\displaystyle {\frac {3}{\sqrt {2x+5}}}}$
(b)     ${\displaystyle y=x+7}$