# Thread: Did i differentiate correctly?

1. ## Did i differentiate correctly?

Wasn't sure if i took the differentiation right!

heres the pic:

https://imgur.com/a/xubrz

2. ## Re: Did i differentiate correctly?

No.

\begin{align*} \dfrac{ d }{ dx } \left( x^{ \tan x } \right)^{ \sec x } & = \dfrac{d}{dx} \left( e^{ \ln x^{ \tan x \sec x } } \right) \\ & = \dfrac{d}{dx} \left( e^{ \tan x \sec x \ln x } \right) \\ & = e^{ \tan x \sec x \ln x } \dfrac{d}{dx} \left( \tan x \sec x \ln x \right) \\ & = x^{ \tan x \sec x } \left( \dfrac{ \sec x \tan x }{ x } + ( \sec^3 x + \sec x \tan^2 x ) \ln x \right) \end{align*}

You had $\sec^2 x+\sec x \tan^2 x$, but it should be $\sec^3 x + \sec x \tan^2 x$

3. ## Re: Did i differentiate correctly?

Originally Posted by SlipEternal
No.

\begin{align*} \dfrac{ d }{ dx } \left( x^{ \tan x } \right)^{ \sec x } & = \dfrac{d}{dx} \left( e^{ \ln x^{ \tan x \sec x } } \right) \\ & = \dfrac{d}{dx} \left( e^{ \tan x \sec x \ln x } \right) \\ & = e^{ \tan x \sec x \ln x } \dfrac{d}{dx} \left( \tan x \sec x \ln x \right) \\ & = x^{ \tan x \sec x } \left( \dfrac{ \sec x \tan x }{ x } + ( \sec^3 x + \sec x \tan^2 x ) \ln x \right) \end{align*}

You had $\sec^2 x+\sec x \tan^2 x$, but it should be $\sec^3 x + \sec x \tan^2 x$

how come you took the e ^ (lnx) instead of just ln of both sides? is it easier to raise it by the exponential or just take the log? Or is it just preference?

4. ## Re: Did i differentiate correctly?

preference, imho ...

$y = x^{\tan{x} \cdot \sec{x}}$

$\ln{y} = \tan{x} \cdot \sec{x} \cdot \ln{x}$

$\dfrac{y'}{y} = \tan{x} \cdot \sec{x} \cdot \dfrac{1}{x} + \tan{x} \cdot \sec{x} \cdot \tan{x} \cdot \ln{x} + \sec^2{x} \cdot \sec{x} \cdot \ln{x}$

$\dfrac{y'}{y} = \sec{x} \bigg[\dfrac{\tan{x}}{x} + \ln{x} \left(2\tan^2{x}+1\right)\bigg]$

$y' = x^{\tan{x} \cdot \sec{x}} \cdot \sec{x} \bigg[\dfrac{\tan{x}}{x} + \ln{x} \left(2\tan^2{x}+1\right)\bigg]$

5. ## Re: Did i differentiate correctly?

Originally Posted by lc99
how come you took the e ^ (lnx) instead of just ln of both sides? is it easier to raise it by the exponential or just take the log? Or is it just preference?
Originally Posted by skeeter
preference, imho ...
$\ln{y} = \tan{x} \cdot \sec{x} \cdot \ln{x}$
Because $\left(x^{\tan(x)}\right)^{\sec(x)}=\exp\left(\tan (x)\sec(x)\log(x)\right)$ we have
$\large\frac{d}{{dx}}{\left( {{x^{\tan (x)}}} \right)^{\sec (x)}} = \exp \left( {\tan (x)\sec (x)\log (x)} \right)\cdot\frac{d}{{dx}}\left( {\tan (x)\sec (x)\log (x)} \right)$

This is my preference.