Given that y=v^(2)+5, find (d)/(dv)(5v^(3)-2sin y) in terms of only v. Answer:

Identify Function and Relationship: Identify the function to differentiate and the given relationship between y and v. We need to differentiate 5v^3 - 2sin(y) with respect to v, and we are given that y = v^2 + 5.Apply Chain Rule: Apply the chain rule to differentiate 5v^3 - 2sin(y) with respect to v. The derivative of 5v^3 with respect to v is 15v^2. The derivative of -2sin(y) with respect to v is -2cos(y) * dy/dv, where dy/dv is the derivative of y with respect to v.Differentiate y = v^2 + 5: Differentiate y = v^2 + 5 with respect to v to find dy/dv. The derivative of v^2 with respect to v is 2v. The derivative of a constant, 5, is 0. So, dy/dv = 2v.Substitute dy/dv: Substitute dy/dv into the derivative of -2sin(y) with respect to v. We have -2cos(y) * dy/dv = -2cos(y) * 2v.Combine Derivatives: Combine the derivatives to get the final answer. The derivative of 5v^3 with respect to v is 15v^2. The derivative of -2sin(y) with respect to v is -4v * cos(y). So, the derivative of 5v^3 - 2sin(y) with respect to v is 15v^2 - 4v * cos(y).Replace y in Final Derivative: Replace y with v^2 + 5 in the final derivative expression. Since y = v^2 + 5, we substitute y with v^2 + 5 in the cosine term to express the derivative entirely in terms of v. The final derivative is 15v^2 - 4v * cos(v^2 + 5).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Given that
y=v^(2)+5, find
(d)/(dv)(5v^(3)-2sin y) in terms of only
v.
Answer:

Given that $y=v^{2}+5$ , find $\frac{d}{d v}\left(5 v^{3}-2 \sin y\right)$ in terms of only $v$ . $\newline$ Answer:

View step-by-step help

Home

Math Problems

Algebra 2

Evaluate exponential functions

Full solution

Q. Given that

y=v^{2}+5

, find

\frac{d}{d v}\left(5 v^{3}-2 \sin y\right)

in terms of only

v

\newline

Answer:

Identify Function and Relationship: Identify the function to differentiate and the given relationship between $y$ and $v$ . We need to differentiate $5v^3 - 2\sin(y)$ with respect to $v$ , and we are given that $y = v^2 + 5$ .
Apply Chain Rule: Apply the chain rule to differentiate $5v^3 - 2\sin(y)$ with respect to $v$ . The derivative of $5v^3$ with respect to $v$ is $15v^2$ . The derivative of $-2\sin(y)$ with respect to $v$ is $-2\cos(y) \cdot \frac{dy}{dv}$ , where $\frac{dy}{dv}$ is the derivative of $y$ with respect to $v$ .
Differentiate $y = v^2 + 5$ : Differentiate $y = v^2 + 5$ with respect to $v$ to find $\frac{dy}{dv}$ . The derivative of $v^2$ with respect to $v$ is $2v$ . The derivative of a constant, $5$ , is $0$ . So, $\frac{dy}{dv} = 2v$ .
Substitute $\frac{dy}{dv}$ : Substitute $\frac{dy}{dv}$ into the derivative of $-2\sin(y)$ with respect to $v$ . We have $-2\cos(y) \cdot \frac{dy}{dv} = -2\cos(y) \cdot 2v$ .
Combine Derivatives: Combine the derivatives to get the final answer. $\newline$ The derivative of $5v^3$ with respect to $v$ is $15v^2$ . $\newline$ The derivative of $-2\sin(y)$ with respect to $v$ is $-4v \cdot \cos(y)$ . $\newline$ So, the derivative of $5v^3 - 2\sin(y)$ with respect to $v$ is $15v^2 - 4v \cdot \cos(y)$ .
Replace $y$ in Final Derivative: Replace $y$ with $v^2 + 5$ in the final derivative expression. $\newline$ Since $y = v^2 + 5$ , we substitute $y$ with $v^2 + 5$ in the cosine term to express the derivative entirely in terms of $v$ . $\newline$ The final derivative is $15v^2 - 4v \cdot \cos(v^2 + 5)$ .