Difference between revisions of "009A Sample Final 1, Problem 3"

Revision as of 17:46, 14 February 2016

Find the derivatives of the following functions.

a) $f(x)=\ln {\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}$

b) $g(x)=2\sin(4x)+4\tan({\sqrt {1+x^{3}}})$

Foundations:
Review chain rule, quotient rule, and derivatives of trig functions

Solution:

(a)

Step 1:
Using the chain rule, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\\end{array}}$

Step 2:
Now, we need to calculate ${\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}$ .
To do this, we use the Chain Rule. So, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {4x}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {4x}{(x^{2}-1)(x^{2}+1)}}\\&&\\&=&\displaystyle {\frac {4x}{x^{4}-1}}\\\end{array}}$

(b)

Step 1:
Again, we need to use the Chain Rule. We have
$g'(x)=8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\bigg (}{\frac {d}{dx}}{\sqrt {1+x^{3}}}{\bigg )}$ .

Step 2:
We need to calculate ${\frac {d}{dx}}{\sqrt {1+x^{3}}}$ .
We use the Chain Rule again to get
:: ${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\bigg (}{\frac {d}{dx}}{\sqrt {1+x^{3}}}{\bigg )}}\\&&\\&=&\displaystyle {8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\frac {1}{2}}(1+x^{3})^{-{\frac {1}{2}}}3x^{2}}\\&&\\&=&\displaystyle {8\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}}\\\end{array}}$

Final Answer:
(a) $f'(x)={\frac {4x}{x^{4}-1}}$
(b) $g'(x)=8\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}$ .

@@ Line 46: / Line 46: @@
 &&\\
 & = & \displaystyle{\frac{4x}{x^4-1}}\\
 \end{array}</math>
 |}
@@ Line 55: / Line 54: @@
 !Step 1: &nbsp;
 |-
-|
+|Again, we need to use the Chain Rule. We have
 |-
-|
+|<math>g'(x)=8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)</math>.
-|-
-|
 |}
@@ Line 65: / Line 62: @@
 !Step 2: &nbsp;
 |-
-|
+|We need to calculate <math>\frac{d}{dx}\sqrt{1+x^3}</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
 |-
-|
+|We use the Chain Rule again to get
 |-
-|
+|::<math>\begin{array}{rcl}
+\displaystyle{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\
+&&\\
+& = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\frac{1}{2}(1+x^3)^{-\frac{1}{2}}3x^2}\\
+&&\\
+& = & \displaystyle{8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}}\\
+\end{array}</math>
 |}
@@ Line 81: / Line 80: @@
 |'''(a)''' <math>f'(x)=\frac{4x}{x^4-1}</math>
 |-
-|'''(b)'''
+|'''(b)''' <math>g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math>.
 |}
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