How do you find #(dy)/(dx)# given #cos(2y)=sqrt(1-x^2)#?

Question

Christopher Agresta · Accepted Answer

# dy/dx = x/(2 sin(2y)sqrt(1 - x^2)) #

or equivalently:

# dy/dx = 1/(2sqrt(1 - x^2)) #

# \ \ \ \ \ cos(2y)=sqrt(1-x^2) # # :. cos(2y)=(1-x^2)^(1/2) # Differentiating implicitly and applying the chain rule we get: # -sin(2y)(2 dy/dx) = 1/2(1 - x^2)^(-1/2) (-2x) # # 2 dy/dx sin(2y) = x/sqrt(1 - x^2) # So we can rearrange to get; # dy/dx = x/(2 sin(2y)sqrt(1 - x^2)) # We can also get an explicit expression should we need it; Using #sin^2A+cos^2A-=1# we have: # sin^2 2y+cos^2 2y=1 # # :. sin^2 2y+(1-x^2)=1 # # :. sin^2 2y=x^2 # # :. sin 2y=x # So the earlier solution can be written as: # \ \ \ \ \ dy/dx = x/(2xsqrt(1 - x^2)) # # :. dy/dx = 1/(2sqrt(1 - x^2)) #

Christopher Agresta · Answer

undefined To find (dy)/(dx), we'll differentiate both sides of the equation with respect to x using implicit differentiation. After differentiation, we'll solve for (dy)/(dx).

Differentiating cos(2y) with respect to x using the chain rule gives: (-sin(2y))*(2(dy)/(dx)).
Differentiating sqrt(1-x^2) with respect to x gives: (-1/2)*(1-x^2)^(-1/2)*(-2x) = x/sqrt(1-x^2).

Setting the derivatives equal gives: (-sin(2y))*(2(dy)/(dx)) = x/sqrt(1-x^2).
Solving for (dy)/(dx), we get: (dy)/(dx) = x/(2sin(2y)sqrt(1-x^2)).