How do you differentiate sin^2x-sin^2y=x-y-5?

2 Answers
Jun 1, 2017

dy/dx = (1-sin2x)/(1-sin2y)

Explanation:

Write the equation as:

sin^2x -x = sin^2y-y-5

Differentiate both sides with respect to x:

2sinxcosx -1 = (2sinycosy -1)dy/dx

dy/dx = (1-sin2x)/(1-sin2y)

Jun 1, 2017

Answer: y'=(sin(2x)-1)/(sin(2y)-1)

Explanation:

Differentiate sin^2x-sin^2y=x-y-5

First, we take the derivative of both sides, leaving dy/(dx) as y':
d/(dx)sin^2x-sin^2y=d/(dx)x-y-5

2sin(x)cos(x)-2sin(y)cos(y)y'=1-y'

Solve for y' by adding 2y'sin(y)cos(y) to both sides and subtracting 1 from both sides:
2sin(x)cos(x)-1=2y'sin(y)cos(y)-y'

2sin(x)cos(x)-1=y'(2sin(y)cos(y)-1)

y'=(2sin(x)cos(x)-1)/(2sin(y)cos(y)-1)

Note that sin(2theta)=2sin(theta)cos(theta)
So, we can write y' as:
y'=(sin(2x)-1)/(sin(2y)-1)