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

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Answer 1 · 2017-06-01T23:14:13

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

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)#

Answer 2 · 2017-06-01T23:15:26

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

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)#