find \\frac{dy}{dx} for y = \\frac{9}{x} + 5\\sin x.\\frac{d}{dx}(\\frac{9}{x} + 5\\sin x) = \\square
find \\frac{dy}{dx} for y = \\frac{9}{x} + 5\\sin x.\\frac{d}{dx}(\\frac{9}{x} + 5\\sin x) = \\square
Answer
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:
$-\frac{9}{x^{2}}+5\cos x$