Given that x=3y^(3)+1, find (d)/(dy)(2x^(2)-2cos y) in terms of only y. Answer:

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Given that  x=3y^(3)+1, find  (d)/(dy)(2x^(2)-2cos y) in terms of only  y. Answer:

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Express Function in Terms of y: First, we need to express the function 2x^2 - 2cos(y) in terms of y using the given relationship x = 3y^3 + 1. Substitute x = 3y^3 + 1 into the function to get a new function of y.Substitute x into Function: The new function in terms of y is 2(3y^3 + 1)^2 - 2cos(y). Now we need to find the derivative of this new function with respect to y.Find Derivative of New Function: To find the derivative, we will use the chain rule for the term 2(3y^3 + 1)^2 and the basic derivative rule for the term -2cos(y). The derivative of -2cos(y) with respect to y is 2sin(y).Derivative of -2cos(y): Now, let's find the derivative of 2(3y^3 + 1)^2 with respect to y. Let u = 3y^3 + 1, then the term becomes 2u^2. The derivative of 2u^2 with respect to u is 4u.Find Derivative of 2(3y^3 + 1)^2: Next, we need to find du/dy, which is the derivative of u = 3y^3 + 1 with respect to y. The derivative of 3y^3 with respect to y is 9y^2, and the derivative of 1 with respect to y is 0. So, du/dy = 9y^2.Find du/dy: Now, we apply the chain rule. The derivative of 2u^2 with respect to y is 4u * du/dy. Substitute u = 3y^3 + 1 and du/dy = 9y^2 into the equation to get 4(3y^3 + 1)(9y^2).Apply Chain Rule: Simplify the expression 4(3y^3 + 1)(9y^2) to get 36y^2(3y^3 + 1).Simplify Expression: Combine the derivatives of both terms to get the final derivative of the function with respect to y. The final derivative is 36y^2(3y^3 + 1) + 2sin(y).

Given that $x=3 y^{3}+1$ , find $\frac{d}{d y}\left(2 x^{2}-2 \cos y\right)$ in terms of only $y$ . $\newline$ Answer:

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Given that x=3y3+1 x=3 y^{3}+1 x=3y3+1, find ddy(2x2−2cos⁡y) \frac{d}{d y}\left(2 x^{2}-2 \cos y\right) dyd​(2x2−2cosy) in terms of only y y y.\newlineAnswer:

Given that $x=3 y^{3}+1$ , find $\frac{d}{d y}\left(2 x^{2}-2 \cos y\right)$ in terms of only $y$ . $\newline$ Answer: