Question

question find the derivative of y = sin⁻¹(7x - 9). be sure to include parentheses around the arguments of any logarithmic or trigonometric functions in your answer. provide your answer below:

Explanation:

Step1: Apply chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Here $f(u)=\sin(u)$ and $u = g(x)=7x - 9$.
The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$, and the derivative of $u = 7x-9$ with respect to $x$ is $7$.

Step2: Substitute back $u$

Substitute $u = 7x - 9$ into the derivative.
$y'=\cos(7x - 9)\cdot7$

Answer:

$y' = 7\cos(7x - 9)$