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Given that 
w=y^(3)+5, find 
(d)/(dy)(5w^(4)+4cos y) in terms of only 
y.
Answer:

Given that w=y3+5 w=y^{3}+5 , find ddy(5w4+4cosy) \frac{d}{d y}\left(5 w^{4}+4 \cos y\right) in terms of only y y .\newlineAnswer:

Full solution

Q. Given that w=y3+5 w=y^{3}+5 , find ddy(5w4+4cosy) \frac{d}{d y}\left(5 w^{4}+4 \cos y\right) in terms of only y y .\newlineAnswer:
  1. Given function and task: Given the function w=y3+5w = y^3 + 5, we need to find the derivative of 5w4+4cos(y)5w^4 + 4\cos(y) with respect to yy. We will use the chain rule for differentiation, which 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.
  2. Find derivative of ww: First, let's find the derivative of ww with respect to yy. Since w=y3+5w = y^3 + 5, we have:\newlinedwdy=d(y3+5)dy\frac{dw}{dy} = \frac{d(y^3 + 5)}{dy}\newlinedwdy=3y2+0\frac{dw}{dy} = 3y^2 + 0\newlinedwdy=3y2\frac{dw}{dy} = 3y^2
  3. Differentiate 5w45w^4: Next, we need to differentiate the function 5w45w^4 with respect to ww and then multiply by the derivative of ww with respect to yy. This is an application of the chain rule.\newlined(5w4)dw=20w3\frac{d(5w^4)}{dw} = 20w^3\newlineThen, we multiply by dwdy\frac{dw}{dy}:\newlined(5w4)dy=20w3×dwdy\frac{d(5w^4)}{dy} = 20w^3 \times \frac{dw}{dy}\newlined(5w4)dy=20w3×3y2\frac{d(5w^4)}{dy} = 20w^3 \times 3y^2
  4. Substitute ww in terms of yy: Now, we need to substitute ww back in terms of yy to express the derivative in terms of yy only.w=y3+5w = y^3 + 5d(5w4)dy=20(y3+5)33y2\frac{d(5w^4)}{dy} = 20(y^3 + 5)^3 \cdot 3y^2
  5. Differentiate 4cos(y)4\cos(y): We also need to differentiate the term 4cos(y)4\cos(y) with respect to yy.d(4cos(y))dy=4sin(y)\frac{d(4\cos(y))}{dy} = -4\sin(y)
  6. Add derivatives: Finally, we add the derivatives of the two terms to find the derivative of the entire function with respect to yy.
    ddy(5w4+4cos(y))=ddy(5w4)+ddy(4cos(y))\frac{d}{dy}(5w^4 + 4\cos(y)) = \frac{d}{dy}(5w^4) + \frac{d}{dy}(4\cos(y))
    ddy(5w4+4cos(y))=20(y3+5)33y24sin(y)\frac{d}{dy}(5w^4 + 4\cos(y)) = 20(y^3 + 5)^3 \cdot 3y^2 - 4\sin(y)
  7. Simplify final answer: Simplify the expression to get the final answer. d(5w4+4cos(y))dy=60y2(y3+5)34sin(y)\frac{d(5w^4 + 4\cos(y))}{dy} = 60y^2(y^3 + 5)^3 - 4\sin(y)

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