find the derivative of the function. y = 9^sin(πx) y = need help? read it watch it submit answer

Explanation:

Step1: Recall chain - rule

If $y = a^{u}$ where $a>0,a\neq1$ and $u$ is a function of $x$, then $y^\prime=a^{u}\ln(a)\cdot u^\prime$. Here $a = 9$, $u=\sin(\pi x)$.

Step2: Find derivative of $u$

If $u=\sin(\pi x)$, by the chain - rule again, if $u = \sin(v)$ and $v=\pi x$, then $u^\prime=\cos(v)\cdot v^\prime$. Since $v=\pi x$, $v^\prime=\pi$. So $u^\prime=\pi\cos(\pi x)$.

Step3: Calculate $y^\prime$

$y^\prime = 9^{\sin(\pi x)}\ln(9)\cdot\pi\cos(\pi x)=\pi\ln(9)\cdot9^{\sin(\pi x)}\cos(\pi x)$.

Answer:

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