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Thread: Did i differentiate correctly?

  1. #1
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    Did i differentiate correctly?

    Wasn't sure if i took the differentiation right!

    heres the pic:

    https://imgur.com/a/xubrz
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  2. #2
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    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
    Last edited by SlipEternal; Nov 10th 2017 at 12:39 PM.
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  3. #3
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    Re: Did i differentiate correctly?

    Quote Originally Posted by SlipEternal View Post
    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?
    Last edited by lc99; Nov 10th 2017 at 02:11 PM.
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  4. #4
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    Re: Did i differentiate correctly?

    preference, imho ...


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

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

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

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

    $y' = x^{\tan{x} \cdot \sec{x}} \cdot \sec{x} \bigg[\dfrac{\tan{x}}{x} + \ln{x} \left(2\tan^2{x}+1\right)\bigg]$
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  5. #5
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    Re: Did i differentiate correctly?

    Quote Originally Posted by lc99 View Post
    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?
    Quote Originally Posted by skeeter View Post
    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.
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