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

3 Answers
Jul 21, 2017

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)

Jul 21, 2017

(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

Jul 21, 2017

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)