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)#