# Radical additive expression in relation to partial derivative.

I am learning the partial derivative and I am stuck on how to proceed with the following expression.

z = √(4 - x^2 - y^2)

the expressions I need to find are Δ_x z,Δ_y z and Δz, which are are intersections in the plane x,y,z, where
Δ_x z = f(x+Δx,y) - f(x,y) Δ_y z = f(x,y+Δy) - f(x,y) Δz = f(x+Δx,y+Δy) - f(x,y)

I have been out of school for a few years now and I can't find any explanations on how to simplify addition expression in a radical such as the expresion z above and this is blocking me.

My first instinct would be to leave the expression complete (for example, Δ_x z = √(4 - (x+Δx)^2 - y^2) - √(4 - x^2 - y^2), but in my previous examples, the  -f(x,y) part of the expressions would simplify the non-delta expressions and that confuses me.

Thank you for your help.

## 1 Answer

If you are trying to compute the partial derivative wrt $x$ of $z = \sqrt{4-x^2-y^2}$, given by $$\frac{\partial z}{\partial x} = \lim_{\Delta x \rightarrow 0} \frac{\sqrt{4-(x+\Delta x)^2-y^2} - \sqrt{4-x^2-y^2}}{\Delta x}$$, a good trick is to multiply numerator and denominator by the conjugate expression $\sqrt{4-(x+\Delta x)^2-y^2} + \sqrt{4-x^2-y^2}$ and simplify. This will eliminate the square roots.

• Unless I am missing something, to multiply by the conjugate expression, I need to multiply it on denominator and numerator (to multiply by 1), so I will be left with the conjugate at the denominator, no? edit : I misread a little bit here. But still, the conjugate expression will still be present at the denominator and I am not sure how to go from there neither. – Jeph Gagnon Apr 30 '18 at 18:26
• After multiplying numerator and denominator, you can distribute and should be able to cancel out a $\Delta x$. – D.B. Apr 30 '18 at 18:40
• I will still be stuck with ${2x}{/}{(}{\sqrt{4-(x+\Delta x)^2-y^2} +\sqrt{4-x^2-y^2}}{)}$ and I still have the same issue as when I started. edit: sorry for the edits, begginer mathjax – Jeph Gagnon Apr 30 '18 at 19:19
• Exactly ! In the limit as $\Delta x$ goes to zero, this becomes $\frac{x}{\sqrt{4-x^2-y^2}}$ – D.B. Apr 30 '18 at 21:24
• which is the answer – D.B. Apr 30 '18 at 21:25