Question

Help with partial derivative.
Partial derivative with respect to x of

Answer 1

\[dz/dx = \frac{ x }{ \sqrt{x ^{2}+y ^{2}} }\]

Answer 2

Are you sure that is \(y^2\) in the denominator and not \(z^2\) ??

Answer 3

It is y^2. The original\[z=\sqrt{x ^{2}+y ^{2}}\] equation calls for

Answer 4

So, this is your original equation and you want to partial derivate it with respect to \(x\) ??

Answer 5

Looking to find the second partial derivative with respect to x. The first partial is \[\frac{ x }{ \sqrt{x ^{2}+y ^{2}} }\]

Answer 6

Yes

Answer 7

This looks good, so we do one more time..

Answer 8

you should use here : Divide Rule of Differentiation..

Answer 9

\[d(\frac{u}{v}) = \frac{v \cdot u' - u \cdot v'}{v^2}\]

Answer 10

I have numbers not working not sure of mistake

Answer 11

\[ \implies \frac{\sqrt{x^2 + y^2} \cdot 1 - x \cdot \frac{x}{\sqrt{x^2 + y^2}}}{x^2+y^2}\]

Answer 12

This step are you getting?

Answer 13

That helps. Let me see what I did and compare. Thank you.

Answer 14

After this, just simplification is there.. Good Luck. :)

Answer 15

I always prefer to use product rule like
\[z=x(x^2+y^2)^{-1/2}\\
\frac{\partial z}{\partial x}=(x^2+y^2)^{-1/2}-x^2(x^2+y^2)^{-3/2 }\]

Answer 16

They are one and the same thing.. :P