Confusing Implicit Differentiation Problem

[tmt1]

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]

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]

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...

 

[MarkFL]

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

Nice...I didn't catch that even though the numerator in the expression I derived is clearly zero via the original equation. (Smirk)