1. Differentiating Inverse Functions

Hey guys I'm having problems with this question, could you guys help me out?

The question is:
Show that $\dfrac{d}{dx}\left(\tan^{-1}(\frac{3}{2}\tan x)\right)=\dfrac{6}{5\sin^2 x+4}$

heres what I did
$\dfrac{d}{dx}\left(\tan^{-1}(\frac{3}{2}\tan x)\right)$

$=\dfrac{\dfrac{3}{2}\times\dfrac{1}{1+x^2}}{1+\lef t(\dfrac{3}{2}\tan x\right)^2}$

$=\dfrac{\dfrac{3}{2(1+x^2)}}{1+\dfrac{9sin^2x}{4co s^2x}}$

$
=\dfrac{\dfrac{3}{2(1+x^2)}}{\dfrac{4cos^2x+9sin^2 x}{4cos^2x}}$

$=\dfrac{6cos^2x}{(1+x^2)(4cos^2x+9sin^2x)}$

$
=\dfrac{6cos^2x}{(1+x^2)(5sin^2x+4)}$

this kinda seems quite close to the result that I had to prove, did I do something wrong?

Thanks

2. Originally Posted by aonin
$=\dfrac{\dfrac{3}{2}\times\dfrac{1}{1+x^2}}{1+\lef t(\dfrac{3}{2}\tan x\right)^2}$

It should be:

$\dfrac{\dfrac{3}{2}\dfrac{1}{\cos^2 x}}{1+\left(\dfrac{3}{2}\tan x\right)^2}$

3. If $\displaystyle y = \arctan{\left(\frac{3}{2}\tan{x}\right)}$

then $\displaystyle \tan{y} = \frac{3}{2}\tan{x}$

$\displaystyle \frac{d}{dx}(\tan{y}) = \frac{d}{dx}\left(\frac{3}{2}\tan{x}\right)$

$\displaystyle \frac{d}{dy}(\tan{y})\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}$

$\displaystyle \sec^2{y}\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}$

$\displaystyle (1 + \tan^2{y})\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}$

$\displaystyle \left\{1 + \tan^2{\left[\arctan{\left(\frac{3}{2}\tan{x}\right)}\right]\right\}\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}$

$\displaystyle \left[1 + \left(\frac{3}{2}\tan{x}\right)^2\right]\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}$

$\displaystyle \frac{dy}{dx} = \frac{\frac{3}{2}\sec^2{x}}{1 + \left(\frac{3}{2}\tan{x}\right)^2}$

4. Originally Posted by FernandoRevilla
It should be:

$\dfrac{\dfrac{3}{2}\dfrac{1}{\cos^2 x}}{1+\left(\dfrac{3}{2}\tan x\right)^2}$
do you mind explaining this step, I didn't really get it

when doing the question I used this formula
$\dfrac{d}{dx}tan^{-1}f(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}$

so wouldn't $f'(x)$ be $u'v+v'u=0\times tanx+\dfrac{1}{1+x^2}\times \dfrac{3}{2}=\dfrac{3}{2(1+x^2)}$

or am I misunderstanding something?

Thanks

5. Originally Posted by aonin
when doing the question I used this formula
$\dfrac{d}{dx}tan^{-1}f(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}$

All right, and $f(x)=(3/2) \tan x$ so, $f'(x)=(3/2)(\tan x)'=(3/2)\sec ^2 x$ .

6. Originally Posted by aonin
do you mind explaining this step, I didn't really get it

when doing the question I used this formula
$\dfrac{d}{dx}tan^{-1}f(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}$

so wouldn't $f'(x)$ be $u'v+v'u=0\times tanx+\dfrac{1}{1+x^2}\times \dfrac{3}{2}=\dfrac{3}{2(1+x^2)}$
You differentiated $tan^{-1}(x)$ again, not $tan(x)$.

or am I misunderstanding something?

Thanks

7. alright I get it now, careless mistake

Thanks a lot

8. Keep this in your mind:

$\displaystyle \frac{d}{dx} \left( arctan(f(x)) \right) = \frac{f'(x)}{1+\left[ f(x) \right]^2}$