What is the derivative of #2^sin(pi*x)#?

2 Answers
Sep 27, 2015

Answer:

#d/dx2^(sin(pix))=2^(sin(pix))*ln2*cospix*(pi)#

Explanation:

Using the following standard rules of differentiation:

#d/dxa^(u(x))=a^u*lna*(du)/dx#

#d/dx sinu(x) = cosu(x)*(du)/dx#

#d/dxax^n=nax^(n-1)#

We obtain the following result:

#d/dx2^(sin(pix))=2^(sin(pix))*ln2*cospix*(pi)#

Sep 27, 2015

Recall that:

#d/(dx)[a^(u(x))] = a^u lna (du)/(dx)#

Thus, you get:

#d/(dx)[2^(sin(pix))]#

#= 2^(sin(pix))*ln2*[cos(pix)*pi]#

#= color(blue)(2^(sin(pix))ln2*picos(pix))#

That means two chain rules. Once on #sin(pix)# and once on #pix#.