009A Sample Midterm 1, Problem 4

Find the derivatives of the following functions. Do not simplify.

a) $f(x)={\sqrt {x}}(x^{2}+2)$

b) $g(x)={\frac {x+3}{x^{\frac {3}{2}}+2}}$ where $x>0$

c) $h(x)={\frac {e^{-5x^{3}}}{\sqrt {x^{2}+1}}}$

Foundations:
1. Product Rule
2. Quotient Rule
3. Chain Rule

Solution:

(a)

Step 1:
Using the Product Rule, we have
$f'(x)=({\sqrt {x}})'(x^{2}+2)+{\sqrt {x}}(x^{2}+2)'.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {({\sqrt {x}})'(x^{2}+2)+{\sqrt {x}}(x^{2}+2)'}\\&&\\&=&\displaystyle {{\bigg (}{\frac {1}{2}}x^{-{\frac {1}{2}}}{\bigg )}(x^{2}+2)+{\sqrt {x}}(2x).}\end{array}}$

(b)

Step 1:
Using the Quotient Rule, we have
$g'(x)={\frac {(x^{\frac {3}{2}}+2)(x+3)'-(x+3)(x^{\frac {3}{2}}+2)'}{(x^{\frac {3}{2}}+2)^{2}}}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\frac {(x^{\frac {3}{2}}+2)(x+3)'-(x+3)(x^{\frac {3}{2}}+2)'}{(x^{\frac {3}{2}}+2)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{\frac {3}{2}}+2)(1)-(x+3)({\frac {3}{2}}x^{\frac {1}{2}})}{(x^{\frac {3}{2}}+2)^{2}}}.}\end{array}}$

(c)

Step 1:

Step 2:

Final Answer:
(a)     ${\bigg (}{\frac {1}{2}}x^{-{\frac {1}{2}}}{\bigg )}(x^{2}+2)+{\sqrt {x}}(2x)$
(b)     ${\frac {(x^{\frac {3}{2}}+2)(1)-(x+3)({\frac {3}{2}}x^{\frac {1}{2}})}{(x^{\frac {3}{2}}+2)^{2}}}$
(c)

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