1. ## Differentiation

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.

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

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

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

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

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

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

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

2. Originally Posted by cozza
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.

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

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

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

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

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

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

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

3. Originally Posted by Chris L T521
I disagree with something here...if $f(x)=\dfrac{e^{\tan x}}{x^3+1}$, then by your method, $e(x)=e^{\tan x}$ and $g(x)=x^3+1$ (not $(x^3+1)^2$). As a result, you're quotient rule should give you $f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}$, which really can't be simplified further....

Sorry, the original question shoul have said $(x^3+1)^2$, not $(x^3+1)$

4. Originally Posted by cozza
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.

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

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

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

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

where did the factor "x" come from?

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

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

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

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

You can use the product rule...

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

then clean that up..