Question

derivative of $e^x$
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$\frac{d}{dx}7cos(x)+3e^x=$

Explanation:

Step1: Apply sum - rule of derivatives

The derivative of a sum $y = u + v$ is $y'=u'+v'$. Let $u = 7\cos(x)$ and $v = 3e^{x}$. Then $\frac{d}{dx}[7\cos(x)+3e^{x}]=\frac{d}{dx}[7\cos(x)]+\frac{d}{dx}[3e^{x}]$.

Step2: Apply constant - multiple rule

The constant - multiple rule states that $\frac{d}{dx}[cf(x)]=c\frac{d}{dx}[f(x)]$. For $\frac{d}{dx}[7\cos(x)]$, $c = 7$ and $f(x)=\cos(x)$, and for $\frac{d}{dx}[3e^{x}]$, $c = 3$ and $f(x)=e^{x}$. So $\frac{d}{dx}[7\cos(x)]+\frac{d}{dx}[3e^{x}]=7\frac{d}{dx}[\cos(x)]+3\frac{d}{dx}[e^{x}]$.

Step3: Use known derivatives

We know that $\frac{d}{dx}[\cos(x)]=-\sin(x)$ and $\frac{d}{dx}[e^{x}]=e^{x}$. Then $7\frac{d}{dx}[\cos(x)]+3\frac{d}{dx}[e^{x}]=7(-\sin(x))+3e^{x}$.

Step4: Simplify

$7(-\sin(x))+3e^{x}=- 7\sin(x)+3e^{x}$.

Answer:

$-7\sin(x)+3e^{x}$