Differentiation - Show steps!

Question

Suppose y is defind implicitly as a function of x by the following equations, where g is a given differentiable function of one variable. Find an expression for y'.

 

a. xy=g(x)+y³

Ans: y' = (g'(x)-y)/(x-3y²)

 

b.g(x+y) = x²+y²

Ans: y' = (2x-g'(x+y))/(g'(x+y)-2y)

 

c. (xy+1)² = g(x²y)

Ans: y' = (2y(xg'(x²y)-xy-1))/(x(2xy+2-xg'(x²y)))

 

 

Answer 1

All of these problems can be solved by implicit differentiation. 

 

In the case of b),   one differentiates the equation with respect to x resulting in:

 

  g'(x+y) (1 + y') =  2x + 2 y y'                 {Notice how the chain rule is used here on the left hand side}

 

This equation can then be solved for y'

 

 [g'(x+y) -2y) ] y'  = 2x -g'(x+y)        and y' =  [ 2x -g'(x+y) ]/[g'(x+y) - 2y ]

 

 Problems a) and c) can be solved using the same approach.

Answer 2

a----

xy'+y=g'(x)+3(y^2)y'

y'[x-3(y^2)]=(g'(x)-y)

 

y'= [(g'(x)-y)/(x-3(y^2))]

 

b----

 

g'(x+y)(1+y')=2x+2yy'

 

y'[g'(x+y)-2y]=[2x-g'(x+y)]

 

y'=[(2x-g'(x+y))/(g'(x+y)-2y)]

 

c-----

 

2(xy+1)(xy'+y)=g'((x^2)y)((x^2)y'+2xy)

 

y'[2x(xy+1)-g'((x^2)y)(x^2)]=[2xyg'((x^2)y)-2y(xy+1)]

 

y'=[(2xyg'((x^2)y)-2y(xy+1)]/[2x(xy+1)-g'((x^2)y)(x^2)]