week 6 group project\n1. use the chain rule to find the derivative of each function.\n a. $y=(x^{3}+1)^{5}$\n…

1a.

Explanation:

Step1: Let $u = x^{3}+1$

$y = u^{5}$

Step2: Find $\frac{du}{dx}$ and $\frac{dy}{du}$

$\frac{du}{dx}=3x^{2}$, $\frac{dy}{du} = 5u^{4}$

Step3: Apply chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u = x^{3}+1$ into $\frac{dy}{du}$: $\frac{dy}{dx}=5(x^{3}+1)^{4}\cdot3x^{2}=15x^{2}(x^{3}+1)^{4}$

Answer:

$15x^{2}(x^{3}+1)^{4}$

1b.

Explanation:

Step1: Let $u = 3x^{3}-4$

$y=\sqrt{u}=u^{\frac{1}{2}}$

Step2: Find $\frac{du}{dx}$ and $\frac{dy}{du}$

$\frac{du}{dx}=9x^{2}$, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$

Step3: Apply chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u = 3x^{3}-4$ into $\frac{dy}{du}$: $\frac{dy}{dx}=\frac{1}{2}(3x^{3}-4)^{-\frac{1}{2}}\cdot9x^{2}=\frac{9x^{2}}{2\sqrt{3x^{3}-4}}$

Answer:

$\frac{9x^{2}}{2\sqrt{3x^{3}-4}}$

1c.

Explanation:

Step1: Let $u = x^{4}$

$y=\sin u$

Step2: Find $\frac{du}{dx}$ and $\frac{dy}{du}$

$\frac{du}{dx}=4x^{3}$, $\frac{dy}{du}=\cos u$

Step3: Apply chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u = x^{4}$ into $\frac{dy}{du}$: $\frac{dy}{dx}=\cos(x^{4})\cdot4x^{3}=4x^{3}\cos(x^{4})$

Answer:

$4x^{3}\cos(x^{4})$

1d.

Explanation:

Step1: Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=x^{2}+3$, $v=(2x^{5}+9)^{4}$

Find $u^\prime$ and $v^\prime$. $u^\prime = 2x$ For $v=(2x^{5}+9)^{4}$, let $t = 2x^{5}+9$, then $v = t^{4}$. $\frac{dt}{dx}=10x^{4}$, $\frac{dv}{dt}=4t^{3}$, so $v^\prime=\frac{dv}{dt}\cdot\frac{dt}{dx}=4(2x^{5}+9)^{3}\cdot10x^{4}=40x^{4}(2x^{5}+9)^{3}$

Step2: Apply the product rule

$(uv)^\prime=2x(2x^{5}+9)^{4}+(x^{2}+3)\cdot40x^{4}(2x^{5}+9)^{3}=2x(2x^{5}+9)^{3}(2x^{5}+9 + 20x^{4}(x^{2}+3))=2x(2x^{5}+9)^{3}(2x^{5}+9+20x^{6}+60x^{4})$

Answer:

$2x(2x^{5}+9)^{3}(2x^{5}+9 + 20x^{6}+60x^{4})$

1e.

Explanation:

Step1: Let $u=-x^{2}+2$

$y = u^{\frac{1}{3}}$

Step2: Find $\frac{du}{dx}$ and $\frac{dy}{du}$

$\frac{du}{dx}=-2x$, $\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}$

Step3: Apply chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u=-x^{2}+2$ into $\frac{dy}{du}$: $\frac{dy}{dx}=\frac{1}{3}(-x^{2}+2)^{-\frac{2}{3}}\cdot(-2x)=-\frac{2x}{3\sqrt[3]{(-x^{2}+2)^{2}}}$

Answer:

$-\frac{2x}{3\sqrt[3]{(-x^{2}+2)^{2}}}$

1f.

Explanation:

Step1: Let $u=\sec(5x)$

$y = u^{2}$

Step2: Find $\frac{du}{dx}$ and $\frac{dy}{du}$

For $u=\sec(5x)$, let $t = 5x$, then $u=\sec t$. $\frac{dt}{dx}=5$, $\frac{du}{dt}=\sec t\tan t$, so $\frac{du}{dx}=5\sec(5x)\tan(5x)$; $\frac{dy}{du}=2u$

Step3: Apply chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u=\sec(5x)$ into $\frac{dy}{du}$: $\frac{dy}{dx}=2\sec(5x)\cdot5\sec(5x)\tan(5x)=10\sec^{2}(5x)\tan(5x)$

Answer:

$10\sec^{2}(5x)\tan(5x)$

2.

Explanation:

Step1: Differentiate $x + y^{3}-y = 1$ implicitly with respect to $x$

$1 + 3y^{2}y^\prime-y^\prime=0$

Step2: Solve for $y^\prime$

$y^\prime(3y^{2}-1)=-1$, so $y^\prime=\frac{-1}{3y^{2}-1}$

Step3: When $x = 1$, substitute into the original equation $1 + y^{3}-y = 1$, which gives $y^{3}-y=0$, $y(y - 1)(y + 1)=0$, so $y=0,1,-1$

Step4: Find the slopes at these $y$ - values

When $y = 0$, $y^\prime=-1$; when $y = 1$, $y^\prime=-\frac{1}{2}$; when $y=-1$, $y^\prime=-\frac{1}{2}$

Step5: Use the point - slope form $y - y_{0}=m(x - x_{0})$

For $(x_{0}=1,y_{0}=0)$ and $m=-1$, the equation is $y-0=-1(x - 1)$, i.e., $y=-x + 1$ For $(x_{0}=1,y_{0}=1)$ and $m=-\frac{1}{2}$, the equation is $y - 1=-\frac{1}{2}(x - 1)$, i.e., $y=-\frac{1}{2}x+\frac{3}{2}$ For $(x_{0}=1,y_{0}=-1)$ and $m=-\frac{1}{2}$, the equation is $y + 1=-\frac{1}{2}(x - 1)$, i.e., $y=-\frac{1}{2}x-\frac{1}{2}$

Answer:

$y=-x + 1$, $y=-\frac{1}{2}x+\frac{3}{2}$, $y=-\frac{1}{2}x-\frac{1}{2}$

3.

Explanation:

Step1: Let $y$ be the height of the elevator above the observer's horizontal line of sight, so $y$ is the vertical distance from the observer's horizontal line to the elevator. The horizontal distance $x = 20$ m. Let $\theta$ be the angle of elevation. Then $\tan\theta=\frac{y}{20}$

Differentiate both sides with respect to time $t$: $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$

Step2: When the elevator is 10 m above the ground, $y=10 - 20=-10$ m

$\tan\theta=\frac{-10}{20}=-\frac{1}{2}$, $\sec^{2}\theta=1+\tan^{2}\theta=1+\frac{1}{4}=\frac{5}{4}$, $\frac{dy}{dt}=5$ m/s Substitute into $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$: $\frac{5}{4}\frac{d\theta}{dt}=\frac{1}{20}\times5$, $\frac{d\theta}{dt}=\frac{1}{5}$ rad/s

Step3: When the elevator is 40 m above the ground, $y=40 - 20 = 20$ m

$\tan\theta=\frac{20}{20}=1$, $\sec^{2}\theta=1 + 1=2$, $\frac{dy}{dt}=5$ m/s Substitute into $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$: $2\frac{d\theta}{dt}=\frac{1}{20}\times5$, $\frac{d\theta}{dt}=\frac{1}{8}$ rad/s

Answer:

When the elevator is 10 m above the ground, $\frac{d\theta}{dt}=\frac{1}{5}$ rad/s; when the elevator is 40 m above the ground, $\frac{d\theta}{dt}=\frac{1}{8}$ rad/s