Question

find $\frac{dy}{dx}$ for $y = \frac{3}{x}+4sin x$.
$\frac{d}{dx}(\frac{3}{x}+4sin x)=square$

Explanation:

Step1: Use sum - rule of differentiation

The derivative of a sum $u + v$ is $\frac{d(u + v)}{dx}=\frac{du}{dx}+\frac{dv}{dx}$. Here $u=\frac{3}{x}$ and $v = 4\sin x$. So $\frac{d}{dx}(\frac{3}{x}+4\sin x)=\frac{d}{dx}(\frac{3}{x})+\frac{d}{dx}(4\sin x)$.

Step2: Differentiate $\frac{3}{x}$

Rewrite $\frac{3}{x}$ as $3x^{-1}$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $a = 3$ and $n=-1$, we have $\frac{d}{dx}(3x^{-1})=3\times(-1)x^{-1 - 1}=-3x^{-2}=-\frac{3}{x^{2}}$.

Step3: Differentiate $4\sin x$

Using the constant - multiple rule $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$ and the derivative of $\sin x$ which is $\cos x$. So $\frac{d}{dx}(4\sin x)=4\frac{d}{dx}(\sin x)=4\cos x$.

Step4: Combine the results

$\frac{d}{dx}(\frac{3}{x}+4\sin x)=-\frac{3}{x^{2}}+4\cos x$.

Answer:

$-\frac{3}{x^{2}}+4\cos x$