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Given that
v=4y^(4)-5, find
(d)/(dy)(3y^(3)-5sin v) in terms of only
y.
Answer:

Given that $v=4 y^{4}-5$ , find $\frac{d}{d y}\left(3 y^{3}-5 \sin v\right)$ in terms of only $y$ . $\newline$ Answer:

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Write a formula for an arithmetic sequence

Full solution

Q. Given that

v=4 y^{4}-5

, find

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

in terms of only

y

.

\newline

Answer:

Find Derivative with Chain Rule: First, we need to find the derivative of the function $3y^3 - 5\sin(v)$ with respect to $y$ . To do this, we will use the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, the outer function is $3y^3 - 5\sin(u)$ (where $u$ is a function of $y$ ), and the inner function is $v = 4y^4 - 5$ .
Differentiate Outer Function: We start by differentiating the outer function with respect to $u$ , treating $u$ as the variable. The derivative of $3y^3$ with respect to $u$ is $0$ , since it does not contain the variable $u$ . The derivative of $-5\sin(u)$ with respect to $u$ is $-5\cos(u)$ . So, the derivative of the outer function with respect to $u$ is $-5\cos(u)$ .
Differentiate Inner Function: Next, we need to differentiate the inner function $v = 4y^4 - 5$ with respect to $y$ . The derivative of $4y^4$ with respect to $y$ is $16y^3$ , and the derivative of $-5$ with respect to $y$ is $0$ . Therefore, the derivative of $v$ with respect to $y$ is $16y^3$ .
Apply Chain Rule: Now, we apply the chain rule by multiplying the derivative of the outer function with respect to $u$ by the derivative of the inner function with respect to $y$ . This gives us the derivative of $3y^3 - 5\sin(v)$ with respect to $y$ as $-5\cos(v) \cdot 16y^3$ .
Express Derivative in Terms of y: Finally, we need to express the derivative in terms of $y$ only. Since $v = 4y^4 - 5$ , we substitute $v$ back into our expression for the derivative. This gives us the final derivative as $-5\cos(4y^4 - 5) \times 16y^3$ .

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