1. find dy/dx for each function.\na) y = cos 3x\nb) y = sin(x + π)\nc) y = sin(2x + 3)\nd) y = cos(-πx +…

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = f(g(x))$, then $\frac{dy}{dx}=f^{\prime}(g(x))\cdot g^{\prime}(x)$. Also, $\frac{d}{dx}(\cos u)=-\sin u\cdot\frac{du}{dx}$ and $\frac{d}{dx}(\sin u)=\cos u\cdot\frac{du}{dx}$.

Step2: Solve 1a

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

Step3: Solve 1b

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

Step4: Solve 1c

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

Step5: Solve 1d

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

Step6: Solve 2a

Let $u = 2\theta$, then $y = 4\cos u$. $\frac{du}{d\theta}=2$ and $\frac{dy}{du}=-4\sin u$. By the chain - rule, $\frac{dy}{d\theta}=-4\sin(2\theta)\cdot2=-8\sin(2\theta)$.

Step7: Solve 2b

Let $u = 2\theta-\pi$, then $y = 3\sin u$. $\frac{du}{d\theta}=2$ and $\frac{dy}{du}=3\cos u$. By the chain - rule, $\frac{dy}{d\theta}=3\cos(2\theta-\pi)\cdot2 = 6\cos(2\theta-\pi)=-6\cos(2\theta)$.

Step8: Solve 2c

Let $u=\theta-\frac{\pi}{2}$, then $y=-2\cos u$. $\frac{du}{d\theta}=1$ and $\frac{dy}{du}=2\sin u$. By the chain - rule, $\frac{dy}{d\theta}=2\sin(\theta-\frac{\pi}{2})\cdot1=2\sin(\theta-\frac{\pi}{2})=-2\cos\theta$.

Step9: Solve 2d

Let $u = 4\theta$, then $y=\pi\sin u$. $\frac{du}{d\theta}=4$ and $\frac{dy}{du}=\pi\cos u$. By the chain - rule, $\frac{dy}{d\theta}=\pi\cos(4\theta)\cdot4 = 4\pi\cos(4\theta)$.

Step10: Solve 3a

Rewrite $y=\cos^{2}x=(\cos x)^{2}$. Let $u=\cos x$, then $y = u^{2}$. $\frac{du}{dx}=-\sin x$ and $\frac{dy}{du}=2u$. By the chain - rule, $\frac{dy}{dx}=2\cos x\cdot(-\sin x)=-\sin(2x)$.

Step11: Solve 3b

Rewrite $f(x)=2\sin^{4}x = 2(\sin x)^{4}$. Let $u=\sin x$, then $y = 2u^{4}$. $\frac{du}{dx}=\cos x$ and $\frac{dy}{du}=8u^{3}$. By the chain - rule, $f^{\prime}(x)=8\sin^{3}x\cdot\cos x$.

Step12: Solve 3c

Rewrite $y=\frac{1}{\sin x}=(\sin x)^{-1}$. Let $u=\sin x$, then $y = u^{-1}$. $\frac{du}{dx}=\cos x$ and $\frac{dy}{du}=-u^{-2}$. By the chain - rule, $\frac{dy}{dx}=-\frac{\cos x}{\sin^{2}x}=-\csc x\cot x$.

Step13: Solve 3d

$f(x)=\sin^{3}x-2\cos x$. The derivative of $\sin^{3}x$: Let $u = \sin x$, then $y = u^{3}$, $\frac{du}{dx}=\cos x$, $\frac{dy}{du}=3u^{2}$, so the derivative of $\sin^{3}x$ is $3\sin^{2}x\cos x$. The derivative of $-2\cos x$ is $2\sin x$. So $f^{\prime}(x)=3\sin^{2}x\cos x + 2\sin x=\sin x(3\sin x\cos x + 2)$.

Step14: Solve 4a

Rewrite $y=\sin^{2}(t - 3)=[\sin(t - 3)]^{2}$. Let $u=\sin(t - 3)$ and $v=t - 3$. Then $\frac{dv}{dt}=1$, $\frac{du}{dv}=\cos v$, and $y = u^{2}$, $\frac{dy}{du}=2u$. By the chain - rule, $\frac{dy}{dt}=2\sin(t - 3)\cos(t - 3)$.

Answer:

1a. $\frac{dy}{dx}=-3\sin(3x)$ 1b. $\frac{dy}{dx}=-\cos x$ 1c. $\frac{dy}{dx}=2\cos(2x + 3)$ 1d. $\frac{dy}{dx}=\pi\sin(-\pi x + 3)$ 2a. $\frac{dy}{d\theta}=-8\sin(2\theta)$ 2b. $\frac{dy}{d\theta}=-6\cos(2\theta)$ 2c. $\frac{dy}{d\theta}=-2\cos\theta$ 2d. $\frac{dy}{d\theta}=4\pi\cos(4\theta)$ 3a. $\frac{dy}{dx}=-\sin(2x)$ 3b. $f^{\prime}(x)=8\sin^{3}x\cos x$ 3c. $\frac{dy}{dx}=-\csc x\cot x$ 3d. $f^{\prime}(x)=\sin x(3\sin x\cos x + 2)$ 4a. $\frac{dy}{dt}=2\sin(t - 3)\cos(t - 3)$