Question

find \frac{dy}{dx} for y = \frac{9}{x} + 5\sin x.\frac{d}{dx}(\frac{9}{x} + 5\sin x) = \square

Explanation:

Step1: Differentiate each term separately

Use the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ for $\frac{9}{x}=9x^{-1}$ and the derivative of $\sin x$ is $\cos x$.
For $y = 9x^{-1}+5\sin x$.
The derivative of $9x^{-1}$:
$$\frac{d}{dx}(9x^{-1})=9\times(-1)x^{-1 - 1}=-9x^{-2}=-\frac{9}{x^{2}}$$
The derivative of $5\sin x$:
$$\frac{d}{dx}(5\sin x)=5\cos x$$

Step2: Combine the derivatives

By the sum rule $\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}$, where $u = 9x^{-1}$ and $v = 5\sin x$.
$$\frac{d}{dx}(9x^{-1}+5\sin x)=\frac{d}{dx}(9x^{-1})+\frac{d}{dx}(5\sin x)$$
Substitute the derivatives from Step1:
$$\frac{d}{dx}(9x^{-1}+5\sin x)=-\frac{9}{x^{2}}+5\cos x$$

Answer:

$-\frac{9}{x^{2}}+5\cos x$