Question

How do you differentiate `y=2^(sinpix)`?

Answer 1

`dy/dx=piln2cos(pix)2^(sinpix)`

Explanation:

You must write `2^(sinpix)` in an other way :

`y=2^(sinpix)=e^((sinpix)ln2)`

Now we differentiate :

`dy/dx=e^((sinpix)ln2)(ln2*pi*cospix)=piln2cos(pix)2^(sinpix)`

Answer 2

`(dy)/(dx)=piln2*(2^(sinpix))cospix`

Explanation:

use lograthmic differentiation

`y=2^(sinpix)`

`=>lny=ln2^(sinpix)`

`=>lny=sinpixln2`

differentiate implicitly

`1/y(dy)/(dx)=picospixln2`

`=>(dy)/(dx)=ypicospixln2`

substitue back for `y`

`(dy)/(dx)=2^(sinpix)picospixln2`

tidying up

`(dy)/(dx)=piln2*(2^(sinpix))cospix`

Answer 3

`dy/dx=piln2cos(pix)2^(sin(pix))`

Explanation:

`•color(white)(x)d/dx(a^x)=a^xlna`

`•color(white)(x)d/dx(a^(f(x)))=a^(f(x))lnaxxf'(x)larr" chain rule"`

`y=2^(sinpix)`

`rArrdy/dx=2^(sinpix)ln2xxd/dx(sinpix)`

`color(white)(rArrdy/dx)=2^(sinpix)ln2xxcos(pix)xxd/dx(pix)`

`color(white)(rArrdy/dx)=piln2cos(pix)2^(sinpix)`