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 $pix$ .

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