009A Sample Midterm 1, Problem 5 Detailed Solution

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

Background Information:
1. Product Rule
${\frac {d}{dx}}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)$
2. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$
3. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

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:
Using the Quotient Rule, we have
$h'(x)={\frac {{\sqrt {x^{2}+1}}(e^{-5x^{3}})'-e^{-5x^{3}}({\sqrt {x^{2}+1}})'}{({\sqrt {x^{2}+1}})^{2}}}.$

Step 2:
Now, using the Chain Rule, we have
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{h'(x)} & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}}\\ &&\\ & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-5x^3)'-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(x^2+1)'}{(\sqrt{x^2+1})^2}}\\ &&\\ & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}.} \end{array}}

Final Answer:
(a)     $f'(x)={\bigg (}{\frac {1}{2}}x^{-{\frac {1}{2}}}{\bigg )}(x^{2}+2)+{\sqrt {x}}(2x)$
(b)     $g'(x)={\frac {(x^{\frac {3}{2}}+2)(1)-(x+3)({\frac {3}{2}}x^{\frac {1}{2}})}{(x^{\frac {3}{2}}+2)^{2}}}$
(c)     $h'(x)={\frac {{\sqrt {x^{2}+1}}(e^{-5x^{3}})(-15x^{2})-e^{-5x^{3}}{\frac {1}{2}}(x^{2}+1)^{\frac {-1}{2}}(2x)}{({\sqrt {x^{2}+1}})^{2}}}$

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