How do you differentiate y=(x-y)^2/y?

1 Answer
Feb 27, 2016

dy/dx=(2(x-y)y)/x^2

Explanation:

First, let's think of this as f(x)=(x-y)^2*y^-1 instead, because it is simpler to use the product rule rather than the quotient rule.

Product Rule: y'=dy/dx=g'(x)*h(x)+h'(x)*g(x)

Let g(x)=(x-y)^2. Let's compute for g'(x).

[1]" "g'(x)=2(x-y) by Power Rule. We need to use Chain Rule after.

[2]" "g'(x)=2(x-y)*(1-1*dy/dx)

[3]" "g'(x)=2(x-y)-[2(x-y)dy/dx]

Let h(x)=y^-1. Let's compute for h'(x).

[1]" "h'(x)=-1*y^-2 by Power Rule. We need to use Chain Rule.

[2]" "h'(x)=-y^-2*1*dy/dx

[3]" "h'(x)=-y^-2dy/dx

Now we can solve for dy/dx.

[1]" "dy/dx=g'(x)*h(x)+h'(x)*g(x)

[2]" "dy/dx={2(x-y)-[2(x-y)dy/dx]}*y^-1+[-y^-2dy/dx]"⋅"[(x-y)^2]

[3]" "dy/dx=2y^-1(x-y)-[2(x-y)dy/dx]y^-1-y^-2dy/dx(x-y)^2

[4]" "dy/dx=(2(x-y))/y-[2(x-y)dy/dx]/y-(dy/dx(x-y)^2)/y^2

Subtract both sides of the equation by the terms containing dy/dx.

[5]" "dy/dx+[2(x-y)dy/dx]/y+(dy/dx(x-y)^2)/y^2=(2(x-y))/y

Factor out dy/dx.

[6]" "dy/dx(1+[2(x-y)]/y+((x-y)^2)/y^2)=(2(x-y))/y

[7]" "dy/dx((y^2+2(x-y)y+(x-y)^2)/y^2)=(2(x-y))/y

[8]" "dy/dx=(2(x-y))/y(y^2/[y-(x-y)]^2)

[9]" "color(blue)(dy/dx=(2(x-y)y)/x^2)

Note: Your solution doesn't actually have to be this long. In fact, you can complete this whole problem in maybe two or three lines. I just broke it down into several parts so you know what I'm doing.