Diffrentiate a terribly hard deriviative

[YehiaMedhat]

[math][cosh(sec(lnx))]^{tanh}^-1[/math]

 

[BigBeachBanana]

[math][cosh(sec(lnx))]^{tanh}^-1[/math]

[imath]\large [\cosh(\sec(\ln x))]^{\tanh^{-1}}\\[/imath]

Start with the generalized power rule: [math][u(x)^{v(x)}]'=u(x)^{v(x)}[\ln(u(x))\cdot v(x)]'[/math]

 

[Harry_the_cat]

Are you sure the exponent is correct?

 

[Dr.Peterson]

[math][cosh(sec(lnx))]^{tanh}^-1[/math]

[imath][\cosh(\sec(\ln x))]^{\tanh^{-1}{\color{red}{(x)}}}[/imath]??

Use the comment in #2, or try logarithmic differentiation.

 

[skeeter]

logarithmic differentiation may be a reasonable option …

[imath]y = \bigg[ f[g(u)] \bigg]^v[/imath]

[imath]\dfrac{d}{dx} \bigg[ \ln{y} = v \cdot \ln\bigg( f[g(u)] \bigg) \bigg][/imath]

 

[YehiaMedhat]

Well, that's good
But isn't there any more shortcut ideas?
I do know how to derivative it but I just wanted some tips and tricks that would make the derivative much easier

 

[Otis]

isn't there any more shortcut ideas? … I do know how to derivative it but I just wanted some tips and tricks

You ought to state what you want in your first post. And, if you're looking for something other than what you've already accomplished, then you need to show what you've already done.

Please follow the forum's posting guidelines. Thanks!
[imath]\;[/imath]

 

[Steven G]

Taking ln of both sides before differentiation is the easiest method that I know of.