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$$f(x)=arctan(8x);
f'(x)=?$$

$$f(x)=sin(x)^{cos(x)};
f'(x)=?$$

YehChi  Jan 13, 2015

Best Answer 

 #2
avatar+18835 
+10

I. $$f(x)=arctan(8x);f'(x)=?$$

$$\tan[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ] = 8x \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\(1+\tan^2(\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ) )\times\left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\left[ 1+(8x)^2} \right]\times\left[\textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\boxed{ \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = \frac{8} {1+(8x)^2} } }$$

II. $$f(x)=sin(x)^{cos(x)};f'(x)=?$$

$$y=\sin{(x)}^{ \cos{(x)} } \quad | \quad \ln() \\ \\
\ln{(y)} = \cos{(x)} *\ln{ (\sin{(x)} ) } \quad | \quad \frac{\ d()}{dx} \\ \\
\frac{y'}{y} = [\cos{(x)}]' * \ln{ (\sin{(x)} ) } + \cos{(x)} *[ \ln{ (\sin{(x)} ) } ]' \\\\
y' = y
\left(
[\cos{(x)}]' * \ln{ (\sin{(x)} ) } + \cos{(x)} *[ \ln{ (\sin{(x)} ) } ]'
\right) \\ \\
y' = \sin{(x)}^{ \cos{(x)} }
\left(
-\sin{(x)} * \ln{ (\sin{(x)} ) } + \cos{(x)} * \frac{ \cos{(x)} } {\sin{(x)} }
\right) \\ \\
y' = \sin{(x)}^{ \cos{(x)} }
\left(
-\sin{(x)} * \ln{ (\sin{(x)} ) } + \frac{ \cos^2{(x)} } {\sin{(x)} }
\right) \\ \\$$

heureka  Jan 13, 2015
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3+0 Answers

 #1
avatar+91489 
+5

You are probably encouraged to do this via formula but I can never remember the formula so I will show you the long way.

 

Besides I never do anything the short way - ask anyone       LOL

 

$$\\y=atan(8x)\\\\
8x=tany\\\\
x=\frac{tany}{8}\\\\
\frac{dx}{dy}=\frac{sec^2y}{8}\\\\
\frac{dx}{dy}=\frac{1}{8cos^2y}\\\\
\frac{dy}{dx}=8cos^2y\\\\$$

 

At this point I used this triangle

 

 

$$\\\frac{dy}{dx}=8[cosy]^2\\\\
\frac{dy}{dx}=8[\frac{1}{\sqrt{1+64x^2}}]^2\\\\
\frac{dy}{dx}=\frac{8}{1+64x^2}\\\\$$

Melody  Jan 13, 2015
 #2
avatar+18835 
+10
Best Answer

I. $$f(x)=arctan(8x);f'(x)=?$$

$$\tan[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ] = 8x \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\(1+\tan^2(\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ) )\times\left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\left[ 1+(8x)^2} \right]\times\left[\textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\boxed{ \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = \frac{8} {1+(8x)^2} } }$$

II. $$f(x)=sin(x)^{cos(x)};f'(x)=?$$

$$y=\sin{(x)}^{ \cos{(x)} } \quad | \quad \ln() \\ \\
\ln{(y)} = \cos{(x)} *\ln{ (\sin{(x)} ) } \quad | \quad \frac{\ d()}{dx} \\ \\
\frac{y'}{y} = [\cos{(x)}]' * \ln{ (\sin{(x)} ) } + \cos{(x)} *[ \ln{ (\sin{(x)} ) } ]' \\\\
y' = y
\left(
[\cos{(x)}]' * \ln{ (\sin{(x)} ) } + \cos{(x)} *[ \ln{ (\sin{(x)} ) } ]'
\right) \\ \\
y' = \sin{(x)}^{ \cos{(x)} }
\left(
-\sin{(x)} * \ln{ (\sin{(x)} ) } + \cos{(x)} * \frac{ \cos{(x)} } {\sin{(x)} }
\right) \\ \\
y' = \sin{(x)}^{ \cos{(x)} }
\left(
-\sin{(x)} * \ln{ (\sin{(x)} ) } + \frac{ \cos^2{(x)} } {\sin{(x)} }
\right) \\ \\$$

heureka  Jan 13, 2015
 #3
avatar+91489 
0

Thanks Heureka,  I didn't know where to start with that second one.

I think you have done the first one different from me too - I shall have to take a look :)

Melody  Jan 13, 2015

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