1. find the equation of the tangent line to y = 5cos(x) at x = π/4. 2. let v(x)=xsin(x). find v(x), v(x) and…

1.

Explanation:

Step1: Recall derivative formula for $y = \cos(x)$

The derivative of $y = \cos(x)$ is $y'=-\sin(x)$. The equation of a tangent - line to $y = f(x)$ at $x = a$ is $y - f(a)=f'(a)(x - a)$. For $y = 5\cos(x)$, $f(x)=5\cos(x)$, so $f'(x)=- 5\sin(x)$. When $x=\frac{\pi}{4}$, $f(\frac{\pi}{4})=5\cos(\frac{\pi}{4})=\frac{5\sqrt{2}}{2}$ and $f'(\frac{\pi}{4})=-5\sin(\frac{\pi}{4})=-\frac{5\sqrt{2}}{2}$.

Step2: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Here, $x_1=\frac{\pi}{4}$, $y_1 = \frac{5\sqrt{2}}{2}$, and $m=-\frac{5\sqrt{2}}{2}$. $y-\frac{5\sqrt{2}}{2}=-\frac{5\sqrt{2}}{2}(x - \frac{\pi}{4})$. $y=-\frac{5\sqrt{2}}{2}x+\frac{5\sqrt{2}\pi}{8}+\frac{5\sqrt{2}}{2}$.

Answer:

$y =-\frac{5\sqrt{2}}{2}x+\frac{5\sqrt{2}\pi}{8}+\frac{5\sqrt{2}}{2}$

2.

Explanation:

Step1: Find the first - derivative using the product rule

The product rule states that if $v(x)=u(x)w(x)$, then $v'(x)=u'(x)w(x)+u(x)w'(x)$. For $v(x)=x\sin(x)$, let $u(x)=x$ and $w(x)=\sin(x)$. Then $u'(x) = 1$ and $w'(x)=\cos(x)$. So $v'(x)=\sin(x)+x\cos(x)$.

Step2: Find the second - derivative

Differentiate $v'(x)=\sin(x)+x\cos(x)$ using the sum rule and product rule. The derivative of $\sin(x)$ is $\cos(x)$, and for $x\cos(x)$ using the product rule ($u = x$, $w=\cos(x)$, $u' = 1$, $w'=-\sin(x)$), we get $v''(x)=\cos(x)+\cos(x)-x\sin(x)=2\cos(x)-x\sin(x)$.

Step3: Find the third - derivative

Differentiate $v''(x)=2\cos(x)-x\sin(x)$. The derivative of $2\cos(x)$ is $-2\sin(x)$, and for $-x\sin(x)$ using the product rule ($u=-x$, $w = \sin(x)$, $u'=-1$, $w'=\cos(x)$), we get $v^{(3)}(x)=-2\sin(x)-\sin(x)-x\cos(x)=-3\sin(x)-x\cos(x)$.

Answer:

$v'(x)=\sin(x)+x\cos(x)$, $v''(x)=2\cos(x)-x\sin(x)$, $v^{(3)}(x)=-3\sin(x)-x\cos(x)$

3.

Explanation:

Step1: Recall the quotient rule

The quotient rule states that if $y=\frac{f(\theta)}{g(\theta)}$, then $\frac{dy}{d\theta}=\frac{f'(\theta)g(\theta)-f(\theta)g'(\theta)}{g(\theta)^2}$. For $y = \cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$, let $f(\theta)=\cos(\theta)$ and $g(\theta)=\sin(\theta)$. Then $f'(\theta)=-\sin(\theta)$ and $g'(\theta)=\cos(\theta)$.

Step2: Apply the quotient rule

$\frac{d}{d\theta}\cot(\theta)=\frac{-\sin(\theta)\sin(\theta)-\cos(\theta)\cos(\theta)}{\sin^{2}(\theta)}=\frac{-(\sin^{2}(\theta)+\cos^{2}(\theta))}{\sin^{2}(\theta)}$. Since $\sin^{2}(\theta)+\cos^{2}(\theta) = 1$, we have $\frac{d}{d\theta}\cot(\theta)=-\csc^{2}(\theta)$.

Answer:

The proof is shown above.

4.

A.

Explanation:

Step1: Use the sum rule and product rule

The sum rule states that $\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}$, and the product rule states that $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. For $y=x^{2}\tan(x)+5x\sec(x)$, let $u = x^{2}\tan(x)$ and $v = 5x\sec(x)$. For $u=x^{2}\tan(x)$, using the product rule ($a = x^{2}$, $b=\tan(x)$, $a' = 2x$, $b'=\sec^{2}(x)$), we get $\frac{d}{dx}(x^{2}\tan(x))=2x\tan(x)+x^{2}\sec^{2}(x)$. For $v = 5x\sec(x)$, using the product rule ($a = 5x$, $b=\sec(x)$, $a' = 5$, $b'=\sec(x)\tan(x)$), we get $\frac{d}{dx}(5x\sec(x))=5\sec(x)+5x\sec(x)\tan(x)$. Then $\frac{d}{dx}(x^{2}\tan(x)+5x\sec(x))=2x\tan(x)+x^{2}\sec^{2}(x)+5\sec(x)+5x\sec(x)\tan(x)$.

Answer:

$2x\tan(x)+x^{2}\sec^{2}(x)+5\sec(x)+5x\sec(x)\tan(x)$

B.

Explanation:

Step1: Use the quotient rule

The quotient rule states that if $y=\frac{f(u)}{g(u)}$, then $\frac{dy}{du}=\frac{f'(u)g(u)-f(u)g'(u)}{g(u)^2}$. For $y=\frac{3u^{5/3}+5\cos(u)}{\sin(u)+2u^{3}}$, let $f(u)=3u^{5/3}+5\cos(u)$ and $g(u)=\sin(u)+2u^{3}$. $f'(u)=5u^{2/3}-5\sin(u)$ and $g'(u)=\cos(u)+6u^{2}$.

Step2: Substitute into the quotient rule

$\frac{d}{du}\left(\frac{3u^{5/3}+5\cos(u)}{\sin(u)+2u^{3}}\right)=\frac{(5u^{2/3}-5\sin(u))(\sin(u)+2u^{3})-(3u^{5/3}+5\cos(u))(\cos(u)+6u^{2})}{(\sin(u)+2u^{3})^{2}}$.

Answer:

$\frac{(5u^{2/3}-5\sin(u))(\sin(u)+2u^{3})-(3u^{5/3}+5\cos(u))(\cos(u)+6u^{2})}{(\sin(u)+2u^{3})^{2}}$

C.

Explanation:

Step1: Use the product rule

The product rule states that if $y = a\cdot b$, then $y'=a'b + ab'$. For $y=\sqrt{t}\cdot\sec(t)=t^{1/2}\sec(t)$, let $a=t^{1/2}$ and $b = \sec(t)$. Then $a'=\frac{1}{2}t^{-1/2}$ and $b'=\sec(t)\tan(t)$.

Step2: Calculate the derivative

$\frac{d}{dt}(\sqrt{t}\cdot\sec(t))=\frac{1}{2\sqrt{t}}\sec(t)+\sqrt{t}\sec(t)\tan(t)$.

Answer:

$\frac{1}{2\sqrt{t}}\sec(t)+\sqrt{t}\sec(t)\tan(t)$

D.

Explanation:

Step1: Use the product rule

The product rule states that if $y = a\cdot b\cdot c$, then $y'=a'bc+ab'c + abc'$. For $y = e^{\theta}\sin(\theta)\cos(\theta)$, let $a = e^{\theta}$, $b=\sin(\theta)$, and $c=\cos(\theta)$. $a'=e^{\theta}$, $b'=\cos(\theta)$, $c'=-\sin(\theta)$.

Step2: Calculate the derivative

$\frac{d}{d\theta}(e^{\theta}\sin(\theta)\cos(\theta))=e^{\theta}\sin(\theta)\cos(\theta)+e^{\theta}\cos(\theta)\cos(\theta)-e^{\theta}\sin(\theta)\sin(\theta)=e^{\theta}(\sin(\theta)\cos(\theta)+\cos^{2}(\theta)-\sin^{2}(\theta))$.

Answer:

$e^{\theta}(\sin(\theta)\cos(\theta)+\cos^{2}(\theta)-\sin^{2}(\theta))$