What is the derivative of $2^{sin (π \cdot x)}$ $2^sin(pi*x)$ ?

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Answer 1 · 2015-09-27T17:40:28

$\frac{d}{d x} 2^{sin (π x)} = 2^{sin (π x)} \cdot ln 2 \cdot cos π x \cdot (π)$ $d/dx2^(sin(pix))=2^(sin(pix))*ln2*cospix*(pi)$

Using the following standard rules of differentiation:

$\frac{d}{d x} a^{u (x)} = a^{u} \cdot ln a \cdot \frac{d u}{d x}$ $d/dxa^(u(x))=a^u*lna*(du)/dx$

$\frac{d}{d x} sin u (x) = cos u (x) \cdot \frac{d u}{d x}$ $d/dx sinu(x) = cosu(x)*(du)/dx$

$\frac{d}{d x} a x^{n} = n a x^{n - 1}$ $d/dxax^n=nax^(n-1)$

We obtain the following result:

$\frac{d}{d x} 2^{sin (π x)} = 2^{sin (π x)} \cdot ln 2 \cdot cos π x \cdot (π)$ $d/dx2^(sin(pix))=2^(sin(pix))*ln2*cospix*(pi)$

Answer 2 · 2015-09-27T23:21:11

Recall that:

$\frac{d}{d x} [a^{u (x)}] = a^{u} ln a \frac{d u}{d x}$ $d/(dx)[a^(u(x))] = a^u lna (du)/(dx)$

Thus, you get:

$\frac{d}{d x} [2^{sin (π x)}]$ $d/(dx)[2^(sin(pix))]$

$= 2^{sin (π x)} \cdot ln 2 \cdot [cos (π x) \cdot π]$ $= 2^(sin(pix))*ln2*[cos(pix)*pi]$

$= 2^{sin (π x)} ln 2 \cdot π cos (π x)$ $= color(blue)(2^(sin(pix))ln2*picos(pix))$

That means two chain rules. Once on $sin (π x)$ $sin(pix)$ and once on $π x$ $pix$ .

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