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

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

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Apply Chain Rule: We need to find the derivative of the expression 2u^3 - 3sin(y) with respect to u. 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 2u^3 - 3sin(y) and the inner function is y = u^5 - 2.Differentiate 2u^3: First, we differentiate 2u^3 with respect to u, which is straightforward: (d)/(du)(2u^3) = 3 * 2u^(3-1) = 6u^2.Differentiate -3sin(y): Next, we differentiate -3sin(y) with respect to y, and then multiply by the derivative of y with respect to u using the chain rule: (d)/(dy)(-3sin(y)) = -3cos(y), and (d)/(du)(y) = (d)/(du)(u^5 - 2) = 5u^(5-1) = 5u^4. So, (d)/(du)(-3sin(y)) = -3cos(y) * 5u^4.Combine Derivatives: Now, we combine the derivatives of both parts: (d)/(du)(2u^3 - 3sin(y)) = 6u^2 - 3cos(y) * 5u^4.Substitute y into Expression: Since y = u^5 - 2, we can substitute y back into the expression to express everything in terms of u: (d)/(du)(2u^3 - 3sin(y)) = 6u^2 - 3cos(u^5 - 2) * 5u^4.Simplify Expression: Simplify the expression: (d)/(du)(2u^3 - 3sin(y)) = 6u^2 - 15u^4cos(u^5 - 2).

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

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