Question

How do you take the derivative of `tan(4x)^tan(5x)`?

Answer 1

`(dy)/(dx)=y{(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)}`

Explanation:

Let,

`y=tan(4x)^tan(5x)`

Taking natural log. ,both sides

`lny=lntan(4x)^tan(5x)`

`lny=tan(5x)*lntan(4x)`

Diff.w.r.t `x` ,Using Product Rule:

`1/y(dy)/(dx)=tan(5x)d/(dx)(lntan(4x))+lntan(4x)d/(dx)(tan(5x))`

`1/y(dy)/(dx)=tan(5x)*4/tan(4x)sec^2(4x)+lntan(4x)5sec^2(5x)`

#1/y(dy)/(dx)=4tan(5x)cos(4x)/sin(4x)xx1/cos^2(4x)+5sec^2(5x)lnt an(4x)#

`1/y(dy)/(dx)=(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)`

`(dy)/(dx)=y{(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)}`