How do you find the derivative of $y = {sin}^{2} x + 2^{sin x}$ $y=sin^2x+2^sinx$ ?

Question

7643 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T11:43:15

$\frac{d y}{d x} = sin 2 x + 2^{sin x} \cdot cos x \cdot ln 2 .$ $dy/dx=sin2x+2^sinx*cosx*ln2.$

$y = {sin}^{2} x + 2^{sin x} = {(sin x)}^{2} + 2^{sin x} .$ $y=sin^2x+2^sinx=(sinx)^2+2^sinx.$

$:. dy/dx=d/dx{(sinx)^2}+d/dx{2^sinx}......(1).$

Using the Chain Rule,

$d/dx{(sinx)^2}=2*(sinx)^(2-1)*d/dx{sinx},$

$=2sinxcosx=sin2x............................(2).$

Further, we know that, $d/dx{a^x}=(a^x)lna.$

Reusing the Chain Rule,

$d/dx{2^sinx}=(2^sinx)ln2d/dx{sinx},$

$=2^sinx*cosx*ln2..................................................(3).$

From $(1),(2), and, (3),$ we have,

$dy/dx=sin2x+2^sinx*cosx*ln2.$

How do you find the derivative of y=sin^2x+2^sinxy=sin2x+2sinxy=sin^2x+2^sinx?