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

Aug 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin 2 x + {2}^{\sin} x \cdot \cos x \cdot \ln 2.$

#### Explanation:

$y = {\sin}^{2} x + {2}^{\sin} x = {\left(\sin x\right)}^{2} + {2}^{\sin} x .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left\{{\left(\sin x\right)}^{2}\right\} + \frac{d}{\mathrm{dx}} \left\{{2}^{\sin} x\right\} \ldots \ldots \left(1\right) .$

Using the Chain Rule,

$\frac{d}{\mathrm{dx}} \left\{{\left(\sin x\right)}^{2}\right\} = 2 \cdot {\left(\sin x\right)}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left\{\sin x\right\} ,$

$= 2 \sin x \cos x = \sin 2 x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right) .$

Further, we know that, $\frac{d}{\mathrm{dx}} \left\{{a}^{x}\right\} = \left({a}^{x}\right) \ln a .$

Reusing the Chain Rule,

$\frac{d}{\mathrm{dx}} \left\{{2}^{\sin} x\right\} = \left({2}^{\sin} x\right) \ln 2 \frac{d}{\mathrm{dx}} \left\{\sin x\right\} ,$

$= {2}^{\sin} x \cdot \cos x \cdot \ln 2. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right) .$

From $\left(1\right) , \left(2\right) , \mathmr{and} , \left(3\right) ,$ we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin 2 x + {2}^{\sin} x \cdot \cos x \cdot \ln 2.$