Differentiation

Mar 2010
31
0
Hi,

The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

\(\displaystyle f(x)=\frac{e^{tanx}}{(x^3+1)}\)

Rules: quotient rule, product rule, sum/multiple rules

forulma: \(\displaystyle f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}\)

\(\displaystyle e(x)=e^{tanx}\), \(\displaystyle u=tanx\)
\(\displaystyle e'(u)=e^u\), \(\displaystyle \frac{du}{dx}=sec^2x\)
\(\displaystyle \frac{de}{du}=e^u\)
\(\displaystyle \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}\)

\(\displaystyle g(x)=(x^3+1)^2\), \(\displaystyle u=x^3+1\)
\(\displaystyle g'(u)=u^2\), \(\displaystyle \frac{du}{dx}=3x^2\)
\(\displaystyle \frac{dg}{du}=2u\)
\(\displaystyle \frac{dg}{dx}=2(x^3+1)3x^2\)

\(\displaystyle f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}\)
\(\displaystyle f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}\)

I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?) (Worried)
 

Chris L T521

MHF Hall of Fame
May 2008
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2,046
Chicago, IL
Hi,

The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

\(\displaystyle f(x)=\frac{e^{tanx}}{(x^3+1)}\)

Rules: quotient rule, product rule, sum/multiple rules

forulma: \(\displaystyle f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}\)

\(\displaystyle e(x)=e^{tanx}\), \(\displaystyle u=tanx\)
\(\displaystyle e'(u)=e^u\), \(\displaystyle \frac{du}{dx}=sec^2x\)
\(\displaystyle \frac{de}{du}=e^u\)
\(\displaystyle \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}\)

\(\displaystyle g(x)=(x^3+1)^2\), \(\displaystyle u=x^3+1\)
\(\displaystyle g'(u)=u^2\), \(\displaystyle \frac{du}{dx}=3x^2\)
\(\displaystyle \frac{dg}{du}=2u\)
\(\displaystyle \frac{dg}{dx}=2(x^3+1)3x^2\)

\(\displaystyle f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}\)
\(\displaystyle f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}\)

I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?) (Worried)
I disagree with something here...if \(\displaystyle f(x)=\dfrac{e^{\tan x}}{x^3+1}\), then by your method, \(\displaystyle e(x)=e^{\tan x}\) and \(\displaystyle g(x)=x^3+1\) (not \(\displaystyle (x^3+1)^2\)). As a result, you're quotient rule should give you \(\displaystyle f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}\), which really can't be simplified further....
 
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Mar 2010
31
0
I disagree with something here...if \(\displaystyle f(x)=\dfrac{e^{\tan x}}{x^3+1}\), then by your method, \(\displaystyle e(x)=e^{\tan x}\) and \(\displaystyle g(x)=x^3+1\) (not \(\displaystyle (x^3+1)^2\)). As a result, you're quotient rule should give you \(\displaystyle f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}\), which really can't be simplified further....

Sorry, the original question shoul have said \(\displaystyle (x^3+1)^2\), not \(\displaystyle (x^3+1)\)
 
Dec 2009
3,120
1,342
Hi,

The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

\(\displaystyle f(x)=\frac{e^{tanx}}{(x^3+1)}\)

Rules: quotient rule, product rule, sum/multiple rules

forulma: \(\displaystyle f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}\)

\(\displaystyle e(x)=e^{tanx}\), \(\displaystyle u=tanx\)
\(\displaystyle e'(u)=e^u\), \(\displaystyle \frac{du}{dx}=sec^2x\)
\(\displaystyle \frac{de}{du}=e^u\)
\(\displaystyle \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}\)


where did the factor "x" come from?


\(\displaystyle g(x)=(x^3+1)^2\), \(\displaystyle u=x^3+1\)
\(\displaystyle g'(u)=u^2\), \(\displaystyle \frac{du}{dx}=3x^2\)
\(\displaystyle \frac{dg}{du}=2u\)
\(\displaystyle \frac{dg}{dx}=2(x^3+1)3x^2\)

\(\displaystyle f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}\)
\(\displaystyle f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}\)

I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?) (Worried)

\(\displaystyle f(x)=\displaystyle\huge\frac{e^{tanx}}{(x^3+1)^2}=e^{tanx}\frac{1}{(x^3+1)^2}=e^{tanx}\left(x^3+1\right)^{-2}\)

You can use the product rule...

\(\displaystyle f'(x)=\displaystyle\huge\frac{1}{(x^3+1)^2}\left(e^{tanx}sec^2x\right)+e^{tanx}\left(-2\frac{1}{(x^3+1)^3}3x^2\right)\)

then clean that up..
 
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