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

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Answer 1 · 2015-09-27T17:40:28

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

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)#

Answer 2 · 2015-09-27T23:21:11

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#.