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=3\sqrt{2x+5},x\ge 0.}

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

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

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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:
We start by finding the slope of the tangent line to  $f(x)=3{\sqrt {2x+5}}$  at  $(2,9).$
Using the derivative calculated in part (a), the slope is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {\frac {3}{\sqrt {2(2)+5}}}\\&&\\&=&\displaystyle {\frac {3}{\sqrt {9}}}\\&&\\&=&\displaystyle {1.}\end{array}}$

Step 2:
Now, the tangent line to  $f(x)=3{\sqrt {2x+5}}$  at  $(2,9)$
has slope  $m=1$  and passes through the point  $(2,9).$
Hence, the equation of this line is
$y=(x-2)+9.$
If we simplify, we get
$y=x+7.$

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

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