1. -/1 points details my notes 0/50 submiss find the derivative of the function. y = 9sin(πx) y = need help…

Explanation:

Step1: Apply chain - rule

Let $u = \sin(\pi x)$, then $y = 9^{u}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ for $y = 9^{u}$. The derivative of $a^{u}$ with respect to $u$ is $a^{u}\ln a$, so $\frac{dy}{du}=9^{u}\ln 9=9^{\sin(\pi x)}\ln 9$.

Step2: Find $\frac{du}{dx}$

For $u=\sin(\pi x)$, by the chain - rule again (let $t = \pi x$, then $u=\sin t$), $\frac{du}{dt}=\cos t$ and $\frac{dt}{dx}=\pi$. So $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}=\pi\cos(\pi x)$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=9^{\sin(\pi x)}\ln 9\cdot\pi\cos(\pi x)$.

Answer:

$\pi\ln 9\cdot9^{\sin(\pi x)}\cos(\pi x)$