1. find dy/dx for each function. a) y = cos 3x b) y = sin(x + π) c) y = sin(2x + 3) d) y = cos(-πx + 3)

Explanation:

Step1: Recall chain - rule

If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

Step2: Solve part a

Let $u = 3x$, then $y=\cos u$. $\frac{dy}{du}=-\sin u$ and $\frac{du}{dx}=3$. So $\frac{dy}{dx}=-\sin(3x)\cdot3=- 3\sin(3x)$.

Step3: Solve part b

Let $u=x + \pi$, then $y = \sin u$. $\frac{dy}{du}=\cos u$ and $\frac{du}{dx}=1$. So $\frac{dy}{dx}=\cos(x+\pi)\cdot1=\cos(x + \pi)=-\cos x$.

Step4: Solve part c

Let $u = 2x+3$, then $y=\sin u$. $\frac{dy}{du}=\cos u$ and $\frac{du}{dx}=2$. So $\frac{dy}{dx}=\cos(2x + 3)\cdot2=2\cos(2x + 3)$.

Step5: Solve part d

Let $u=-\pi x+3$, then $y=\cos u$. $\frac{dy}{du}=-\sin u$ and $\frac{du}{dx}=-\pi$. So $\frac{dy}{dx}=-\sin(-\pi x + 3)\cdot(-\pi)=\pi\sin(-\pi x + 3)$.

Answer:

a. $\frac{dy}{dx}=-3\sin(3x)$ b. $\frac{dy}{dx}=-\cos x$ c. $\frac{dy}{dx}=2\cos(2x + 3)$ d. $\frac{dy}{dx}=\pi\sin(-\pi x + 3)$