Given that v=4u^(3)+3, find (d)/(du)(3u^(2)-2sin v) in terms of only u. Answer:

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Given that  v=4u^(3)+3, find  (d)/(du)(3u^(2)-2sin v) in terms of only  u. Answer:

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Apply Chain Rule: First, we need to apply the chain rule to differentiate the expression with respect to u. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. In this case, we have an outer function which is 3u^2 - 2sin(v) and an inner function v(u) = 4u^3 + 3.Differentiate Outer Function: We start by differentiating the outer function with respect to v, which is the variable inside the sine function. The derivative of 3u^2 with respect to v is 0, since it does not contain v. The derivative of -2sin(v) with respect to v is -2cos(v).Differentiate Inner Function: Now we differentiate the inner function v = 4u^3 + 3 with respect to u. The derivative of 4u^3 with respect to u is 12u^2, and the derivative of the constant 3 is 0. So, dv/du = 12u^2.Multiply Derivatives: Using the chain rule, we multiply the derivative of the outer function by the derivative of the inner function. This gives us the derivative of the entire expression with respect to u: (0 - 2cos(v)) * (12u^2).Simplify Expression: Simplify the expression by distributing the 12u^2 across the terms inside the parentheses: -2cos(v) * 12u^2 = -24u^2cos(v).Substitute back into Expression: Finally, we need to express the derivative in terms of u only. Since v = 4u^3 + 3, we substitute v back into our expression for the derivative. This gives us -24u^2cos(4u^3 + 3).

Given that $v=4 u^{3}+3$ , find $\frac{d}{d u}\left(3 u^{2}-2 \sin v\right)$ in terms of only $u$ . $\newline$ Answer:

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Given that v=4u3+3 v=4 u^{3}+3 v=4u3+3, find ddu(3u2−2sin⁡v) \frac{d}{d u}\left(3 u^{2}-2 \sin v\right) dud​(3u2−2sinv) in terms of only u u u.\newlineAnswer:

Given that $v=4 u^{3}+3$ , find $\frac{d}{d u}\left(3 u^{2}-2 \sin v\right)$ in terms of only $u$ . $\newline$ Answer: