Question

if $g(z)=sin(z^{2}-9z + 14)$, find $g(z)=$

Explanation:

Step1: Identify the outer - inner functions

Let $u = z^{2}-9z + 14$, then $g(z)=\sin(u)$.

Step2: Differentiate the outer function

The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$. So, $\frac{d}{du}\sin(u)=\cos(u)$.

Step3: Differentiate the inner function

The derivative of $u = z^{2}-9z + 14$ with respect to $z$ is $\frac{du}{dz}=2z - 9$.

Step4: Apply the chain - rule

The chain - rule states that $g^{\prime}(z)=\frac{dg}{du}\cdot\frac{du}{dz}$. Substituting the results from Step 2 and Step 3, we get $g^{\prime}(z)=\cos(z^{2}-9z + 14)\cdot(2z - 9)$.

Answer:

$(2z - 9)\cos(z^{2}-9z + 14)$