What is the implicit derivative of #3=(1-y)/x^2+xy #?

Question

Charles Arocha · Accepted Answer

#y'=(2 - y(2 + x^3))/(-x + x^4)# First, you should take the derivative of on #3=(1-y)/x^2 + xy# #d/dx3=d/dx( 1-y)/x^2 + d/dxxy# Use the Quotient Rule on #( 1-y)/x^2# Quotient Rule: #f(x)/g(x) = (f'(x)g(x) - g'(x)f(x))/(g(x))^2# #(f'(x)g(x) - g'(x)f(x))/(g(x))^2# = #((-y') * x^2 - 2x * (1-y))/(x^2)^2# It is #-y'# because # d/dx (1-y)# Derivative of any constant is 0, and the derivative of #-y#, using #d/dx# is #dy/dx#, which is #-y'#. Use Product Rule for #d/dx x*y# Product Rule: #f(x)*g(x) = f'(x)*g(x) + g'(x)*f(x)# #f'(x)*g(x) + g'(x)*f(x)# = #d/dx x*y + d/dx y *x# #d/dx x*y + d/dx y *x# = #1 *y + y' *x = y +y'x# Now plug in: #d/dx3=d/dx( 1-y)/x^2 + d/dxxy# #0=((-y') * x^2 - 2x * (1-y))/(x^2)^2 + y +y'x# #0=(-x^2y' - 2x+2xy)/(x^4) + y +y'x# #0=(-y')/x^2 - 2/x^3 + (2y)/x^3 + y +y'x# #0=(-y')/x^2 - (2(1 + y))/x^3 + y +y'x# Let's get the denominator to #x^3# #0=(-xy')/x^3 - (2 + 2y)/x^3 + (x^3y)/x^3 +(y'x^4)/x^3# #0=(-xy')/x^3 + (y'x^4)/x^3- (2 + 2y)/x^3 + (x^3y)/x^3# #0=(-xy' + y'x^4- 2 + 2y + x^3y)/x^3# #0=(y'(-x + x^4)- 2 + y(2 + x^3))/x^3# #0=y'(-x + x^4)- 2 + y(2 + x^3)# #2 - y(2 + x^3)=y'(-x + x^4)# #(2 - y(2 + x^3))/(-x + x^4)=y'#

Christopher Allers · Answer

undefined To find the implicit derivative of $ 3 = \frac{{1-y}}{{x^2}} + xy $, differentiate both sides of the equation with respect to $ x $ using the product rule and chain rule where necessary.

$ \frac{{d}}{{dx}}[3] = 0 $

$ \frac{{d}}{{dx}}\left[\frac{{1-y}}{{x^2}}\right] = \frac{{-2(1-y)}}{{x^3}} - \frac{{(1-y)(-2x)}}{{x^4}} $

$ \frac{{d}}{{dx}}[xy] = y + x\frac{{dy}}{{dx}} $

Then, rearrange terms and solve for $ \frac{{dy}}{{dx}} $.