find $\frac{dy}{dx}$\n$y = cos x-7sqrt{x}+\frac{7}{e^{x}}$\n$\frac{dy}{dx}=square$ (type an exact answer.)

Explanation:

Step1: Differentiate $\cos x$

The derivative of $\cos x$ with respect to $x$ is $-\sin x$.

Step2: Differentiate $-7\sqrt{x}$

Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$. Then using the power - rule $\frac{d}{dx}(ax^n)=anx^{n - 1}$, for $a=-7$ and $n=\frac{1}{2}$, we have $\frac{d}{dx}(-7x^{\frac{1}{2}})=-7\times\frac{1}{2}x^{\frac{1}{2}-1}=-\frac{7}{2\sqrt{x}}$.

Step3: Differentiate $\frac{7}{e^x}$

Rewrite $\frac{7}{e^x}$ as $7e^{-x}$. Using the chain - rule $\frac{d}{dx}(ae^{bx})=abe^{bx}$, for $a = 7$ and $b=-1$, we get $\frac{d}{dx}(7e^{-x})=-7e^{-x}$.

Step4: Sum up the derivatives

By the sum - rule of differentiation $\frac{d}{dx}(u + v+w)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}$, where $u = \cos x$, $v=-7\sqrt{x}$, and $w=\frac{7}{e^x}$, we have $\frac{dy}{dx}=-\sin x-\frac{7}{2\sqrt{x}}-7e^{-x}$.

Answer:

$-\sin x-\frac{7}{2\sqrt{x}}-7e^{-x}$