Question

find $\frac{dy}{dx}$. $y = cos x-2sqrt{x}+\frac{8}{e^{x}}$. $\frac{dy}{dx}=square$ (type an exact answer.)

Explanation:

Step1: Differentiate each term separately

We know that the derivative of $\cos x$ is $-\sin x$, the derivative of $2\sqrt{x}=2x^{\frac{1}{2}}$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ is $2\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$, and the derivative of $\frac{8}{e^x}=8e^{-x}$ using the chain - rule $\frac{d}{dx}(e^{ax}) = ae^{ax}$ is $8\times(-e^{-x})=- \frac{8}{e^x}$.

Step2: Combine the derivatives

By the sum - difference rule of differentiation $\frac{d}{dx}(u\pm v\pm w)=\frac{du}{dx}\pm\frac{dv}{dx}\pm\frac{dw}{dx}$, where $u = \cos x$, $v = 2\sqrt{x}$, and $w=\frac{8}{e^x}$. So $\frac{dy}{dx}=-\sin x-\frac{1}{\sqrt{x}}-\frac{8}{e^x}$.

Answer:

$-\sin x-\frac{1}{\sqrt{x}}-\frac{8}{e^x}$