Calculus integration problem

Question

Use implicit differention:

1. find dy/dx if x^2 - 16xy +y^2 = 1
2. find dy/dx if cosy - sinx = siny - cosx
3. find dy/dx if xsiny + ysinx = pi/2√2 at [pi/4, pi/4]
4. find d^2y/dx^2 if x^2 + 4y^2 = 1

Answer 1

Hi! First off, this isn't an integration problem. It's a differentiation problem.

Implicit differentiation is an application of the Chain Rule. Since in each case you're differentiating with respect to x, you are to regard y as dependent on x.

#1:

(d/dx)(x² - 16xy + y²) = (d/dx)(1) Use the Product Rule when differentiating -16xy.

2x - 16y - 16xy' + 2yy' = 0 It's just algebra now. Solve for y'.

2yy' - 16xy' = 16y - 2x

2(y - 8x)y' = 2(8y - x)

y' = (8y - x)/(y - 8x)

#2:

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

-sin(y)y' - cos(x) = cos(y)y' + sin(x) Now solve for y'

-sin(y)y' - cos(y)y' = sin(x) + cos(x)

-(sin(y) + cos(y))y' = sin(x) + cos(x)

y' = -(sin(x) + cos(x))/(sin(y) + cos(y))

#3:

(d/dx)(xsiny + ysinx) = (d/dx)(pi/2√2) Use the Product Rule on each term on the left side.

sin(y) + xcos(y)y' + y'sin(x) + ycos(x) = 0 Solve for y'.

xcos(y)y' + y'sin(x) = -sin(y) - ycos(x)

y'(xcos(y) + sin(x)) = -(sin(y) + ycos(x))

y' = -(sin(y) + ycos(x))/(xcos(y) + sin(x)) Finally, plug in x = π/4 and y = π/4.

y' = -(sin(π/4) + (π/4)cos(π/4))/((π/4)cos(π/4) + sin(π/4))

y' = -1

#4:

(d/dx)(x² + 4y²) = (d/dx)1

2x + 8yy' = 0 Solve for y'.

8yy' = -2x

y' = -x/(4y) Now find y'', using the Quotient Rule on the right side.

y'' = -(4y - 4xy')/(16y²)

y'' = -(y - xy')/(4y²) Plug in y' = -x/(4y)

y'' = -(y - x(-x/(4y)))/(4y²) Multiply top and bottom by 4y to simplify.

y'' = -(4y² + x)/(16y³) Recall from the problem statement that x² + 4y² = 1.

y'' = -1/(16y³)