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   ${\frac {dy}{dx}}.$

(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:
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
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\cos(\cos(e^{x}))(\cos(e^{x}))'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x})'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x}).}\end{array}}$

Final Answer:
(a)     $f'(x)=3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(14x)$
(b)     $g'(x)=\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x})$

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