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Confusing Implicit Differentiation Problem

tmt

Active member
Jan 15, 2014
236
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 26, 2012
1,404
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

Administrator
Staff member
Feb 24, 2012
13,775
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)