Question

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derivative of $e^x$
$\frac{d}{dx}2cos(x)-3ln(x)-1=$

Explanation:

Step1: Apply sum - difference rule of derivatives

The derivative of a sum/difference of functions $y = u - v - w$ is $y'=u' - v' - w'$. Here $u = 2\cos(x)$, $v = 3\ln(x)$ and $w = 1$. So $\frac{d}{dx}[2\cos(x)-3\ln(x)-1]=\frac{d}{dx}(2\cos(x))-\frac{d}{dx}(3\ln(x))-\frac{d}{dx}(1)$.

Step2: Apply constant - multiple rule

The constant - multiple rule states that $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$. So $\frac{d}{dx}(2\cos(x)) = 2\frac{d}{dx}(\cos(x))$, $\frac{d}{dx}(3\ln(x))=3\frac{d}{dx}(\ln(x))$ and $\frac{d}{dx}(1) = 0$.

Step3: Find derivatives of basic functions

We know that $\frac{d}{dx}(\cos(x))=-\sin(x)$ and $\frac{d}{dx}(\ln(x))=\frac{1}{x}$. Then $2\frac{d}{dx}(\cos(x))=2(-\sin(x))=- 2\sin(x)$ and $3\frac{d}{dx}(\ln(x)) = 3\times\frac{1}{x}=\frac{3}{x}$.

Step4: Combine results

$\frac{d}{dx}(2\cos(x))-\frac{d}{dx}(3\ln(x))-\frac{d}{dx}(1)=-2\sin(x)-\frac{3}{x}-0=-2\sin(x)-\frac{3}{x}$.

Answer:

$-2\sin(x)-\frac{3}{x}$