# Confusing Implicit Differentiation Problem

#### tmt

##### Active member
Hi,

I have

x =(x^2+y^2)^[1/2]

I differentiate

1= 1/2 (x^2+y^2)[-1/2] (2x+2yy')

So far so good. I try to multiply this out.

1= (2x)/2 (x^2+y^2)[-1/2] + (2yy'/2)(x^2+y^2)[-1/2]

I solve for y'

y'= 1/{(x (x^2+y^2)[-1/2]} / {y(x^2+y^2)[-1/2] }

1/x (x^2+y^2)[1/2] * 1/y (x^2+y^2)[1/2]

The square roots multiply out.

y'=(x^2+y^2)/xy

yet the correct response is

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

Am I missing something?

Thanks

#### MarkFL

Staff member
I can't follow what you did after attempting to solve for $y'$.

We are given:

$$\displaystyle x=\left(x^2+y^2 \right)^{\frac{1}{2}}$$

Implicitly differentiating, we obtain:

$$\displaystyle 1=\frac{x+yy'}{\left(x^2+y^2 \right)^{\frac{1}{2}}}$$

or:

$$\displaystyle \left(x^2+y^2 \right)^{\frac{1}{2}}=x+yy'$$

Solve for $y'$:

$$\displaystyle y'=\frac{\left(x^2+y^2 \right)^{\frac{1}{2}}-x}{y}$$

#### Prove It

##### Well-known member
MHB Math Helper
Hi,

I have

x =(x^2+y^2)^[1/2]

I differentiate

1= 1/2 (x^2+y^2)[-1/2] (2x+2yy')

So far so good. I try to multiply this out.

1= (2x)/2 (x^2+y^2)[-1/2] + (2yy'/2)(x^2+y^2)[-1/2]

I solve for y'

y'= 1/{(x (x^2+y^2)[-1/2]} / {y(x^2+y^2)[-1/2] }

1/x (x^2+y^2)[1/2] * 1/y (x^2+y^2)[1/2]

The square roots multiply out.

y'=(x^2+y^2)/xy

yet the correct response is

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

Am I missing something?

Thanks
Surely the derivative is 0. Don't believe me, square both sides of your original equation and see what you're left with...