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

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

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Find Derivative with Chain Rule: First, we need to find the derivative of the function 3y^3 - 5sin(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 - 5sin(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 -5sin(u) with respect to u is -5cos(u). So, the derivative of the outer function with respect to u is -5cos(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 - 5sin(v) with respect to y as -5cos(v) * 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 -5cos(4y^4 - 5) * 16y^3.

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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Given that v=4y4−5 v=4 y^{4}-5 v=4y4−5, find ddy(3y3−5sin⁡v) \frac{d}{d y}\left(3 y^{3}-5 \sin v\right) dyd​(3y3−5sinv) in terms of only y y y.\newlineAnswer:

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: