(a) Find y" (0), if y(x) is implicitly defined by x5 + 4y = x + y5 + 3 and y (0) = 1. (b) Given y = (sin x)", find yí

(a) To find y"(0), we will need to differentiate the equation x^5 + 4y = x + y^5 + 3 twice with respect to x and then evaluate at x=0.

Differentiating once, we get:

5x^4 + 4y' = 1 + 5y^4 y'

Differentiating again, we get:

20x^3 + 4y'' = 20y^3 (y')^2 + 20y^3 y''

At x=0, we know y(0)=1, and we can solve for y'(0) by substituting x=0 into the original equation:

0^5 + 4y(0) = 0 + y(0)^5 + 3

4 = 4

Therefore, y'(0) = 0.

To find y"(0), we need to evaluate the expression we found for y'' at x=0 and substitute in y'(0) = 0 and y(0) = 1:

20(0)^3 + 4y"(0) = 20(1)^3 (0)^2 + 20(1)^3 y"(0)

4y"(0) = 0

y"(0) = 0

Therefore, y"(0) = 0.

(b) Given y = (sin x)", we can find y' by differentiating once with respect to x:

y' = (sin x)'

Using the chain rule, we know that (sin x)' = cos x. Therefore, y' = cos x.

To find y", we will differentiate y' with respect to x:

y" = -sin x

Therefore, y' = cos x and y" = -sin x.