Find y" by implicit differentiation.

cos(y) + sin(x) = 1

Answers

Answer 1

y" = cos(y) * dy/dx - sin(x) + sin(y) by implicit differentiation.

To find the second derivative (y") by implicit differentiation, we will differentiate the equation with respect to x twice.

Equation: cos(y) + sin(x) = 1

Differentiating once with respect to x using the chain rule:

-sin(y) * dy/dx + cos(x) = 0

Now, differentiating again with respect to x:

Differentiating the first term:

-d/dx(sin(y)) * dy/dx - sin(y) * d^2y/dx^2

Differentiating the second term:

-d/dx(cos(x)) = -(-sin(x)) = sin(x)

The equation becomes:

-d/dx(sin(y)) * dy/dx - sin(y) * d^2y/dx^2 + sin(x) = 0

Now, let's isolate the second derivative, d^2y/dx^2:

-d^2y/dx^2 = d/dx(sin(y)) * dy/dx - sin(x) + sin(y)

Substituting the previously obtained expression for d/dx(sin(y)) = cos(y):

-d^2y/dx^2 = cos(y) * dy/dx - sin(x) + sin(y)

Thus, the second derivative (y") by the equation:

y" = cos(y) * dy/dx - sin(x) + sin(y)

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