find \\( \\frac { d y } { d x } \\).\n\n\\( y = \\cos x - 7 \\sqrt { x } + \\frac { 4 } { e ^ { x } }…

find \\( \\frac { d y } { d x } \\).\n\n\\( y = \\cos x - 7 \\sqrt { x } + \\frac { 4 } { e ^ { x } } \\)\n\n\\( \\frac { d y } { d x } = \\square \\) (type an exact answer.)

find \\( \\frac { d y } { d x } \\).\n\n\\( y = \\cos x - 7 \\sqrt { x } + \\frac { 4 } { e ^ { x } } \\)\n\n\\( \\frac { d y } { d x } = \\square \\) (type an exact answer.)

Answer

Explanation:

Step1: Differentiate each term

Differentiate $\cos x$, $-7\sqrt{x}$, and $\frac{4}{e^x}$ separately.

Step2: Differentiate $\cos x$

The derivative of $\cos x$ is $-\sin x$.

Step3: Differentiate $-7\sqrt{x}$

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

Step4: Differentiate $\frac{4}{e^x}$

Rewrite $\frac{4}{e^x}$ as $4e^{-x}$. Using the chain - rule $\frac{d}{dx}(ae^{bx})=abe^{bx}$ (here $a = 4$ and $b=-1$), the derivative is $4\times(-1)e^{-x}=-4e^{-x}=-\frac{4}{e^x}$.

Answer:

$-\sin x-\frac{7}{2\sqrt{x}}-\frac{4}{e^x}$