question find the derivative of y = sin^(-7x - 9)(x). be sure to include parentheses around the arguments of…

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Here, $y = \sin^{-7x - 9}(x)$. Let $u=x$ and $v=-7x - 9$, so $y=\sin^{v}(u)$.

Step2: Rewrite the function

We can rewrite $y$ as $y = (\sin(x))^{-7x - 9}$. Then, take the natural logarithm of both sides: $\ln y=(-7x - 9)\ln(\sin(x))$.

Step3: Differentiate both sides

Differentiating the left - hand side with respect to $x$ gives $\frac{y'}{y}$. For the right - hand side, use the product rule $(uv)' = u'v+uv'$, where $u=-7x - 9$ and $v = \ln(\sin(x))$. $u'=-7$ and $v'=\frac{\cos(x)}{\sin(x)}=\cot(x)$. So, $((-7x - 9)\ln(\sin(x)))'=-7\ln(\sin(x))+(-7x - 9)\cot(x)$.

Step4: Solve for $y'$

Multiply both sides by $y = (\sin(x))^{-7x - 9}$ to get $y'=(\sin(x))^{-7x - 9}\left[-7\ln(\sin(x))-(7x + 9)\cot(x)\right]$.

Answer:

$y'=(\sin(x))^{-7x - 9}\left[-7\ln(\sin(x))-(7x + 9)\cot(x)\right]$