find $\frac{dz}{dy}$ for $z = cos^{2}(3y)$

Explanation:

Step1: Use chain - rule

Let $u = \cos(3y)$. Then $z = u^{2}$. By the chain - rule $\frac{dz}{dy}=\frac{dz}{du}\cdot\frac{du}{dy}$.

Step2: Differentiate $z$ with respect to $u$

$\frac{dz}{du}=\frac{d}{du}(u^{2}) = 2u$.

Step3: Differentiate $u$ with respect to $y$

Let $v = 3y$. Then $u=\cos(v)$. First, $\frac{dv}{dy}=3$, and $\frac{du}{dv}=-\sin(v)$. By the chain - rule, $\frac{du}{dy}=\frac{du}{dv}\cdot\frac{dv}{dy}=-\sin(v)\cdot3=- 3\sin(3y)$.

Step4: Calculate $\frac{dz}{dy}$

Substitute $u = \cos(3y)$ and $\frac{du}{dy}=-3\sin(3y)$ into $\frac{dz}{dy}=\frac{dz}{du}\cdot\frac{du}{dy}$. We get $\frac{dz}{dy}=2u\cdot(-3\sin(3y))=2\cos(3y)\cdot(-3\sin(3y))=-6\cos(3y)\sin(3y)$.

Answer:

$-6\cos(3y)\sin(3y)$