Using the definition, compute the partial derivatives

Question

Consider the following function:

$$f : R^3 → R,\space\space f(x, y, z) = \sqrt{(x^2 + y^2 + z^2)^3}$$

Using the definition, compute the partial derivatives of $f$ at $(x, y, z) = (0, 0, 0).$

Solving with respect to $x$

Definition:

$$\frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{f(x+h,y,z)-f(x,y,z)}{h}$$

Substiuting our function $f$ in the definition, we see;

$$\frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{\sqrt{((x+h)^2+y^2+z^2)^3}-\sqrt{(x^2+y^2+z^2})^3}{h}$$

Using the fact that $f(0,0,0)=0,$

$$\frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{\sqrt{((0+h)^2+0+0)^3}-0}{h}=\lim_{h\to 0}\left(\frac{h^3}{h}\right)=\lim_{h\to 0}h^2=0$$

We would then find the same for $y$ and $z$.

Is my solution correct?

Answer 1

It's almost correct. The definition of derivaitve includes a limit:

$$\frac{\partial f}{\partial x}= \lim_{h \to 0}\frac{f(x+h,y,z)-f(x,y,z)}{h}$$

Perform now the needed calculations and take the limit. Of course, the quantity $\frac{\partial f}{\partial x}$ must not depend on $h$

Answer 2

Denote: $x^2+y^2+z^2=t$ for simplicity of calculations.

The partial derivative is:

\begin{align} \frac{\partial f}{\partial x} &= \lim_\limits{h\to 0}\frac{\sqrt{((x+h)^2+y^2+z^2)^3}-\sqrt{(x^2+y^2+z^2)^3}}{h} \\ &=\lim_\limits{h\to 0}\frac{((x+h)^2+y^2+z^2)^3-(x^2+y^2+z^2)^3}{h[\sqrt{((x+h)^2+y^2+z^2)^3}+\sqrt{(x^2+y^2+z^2)^3}]} \\ &\stackrel{*}=\lim_\limits{h\to 0}\frac{(2xh+h^2)[2xh+h^2+t)^2+(2xh+h^2+t)t+t^2]}{h(\sqrt{(2xh+h^2+t)^3}+\sqrt{t^3})} \\ &=\lim_\limits{h\to 0}\frac{(2x+h)[2xh+h^2+t)^2+(2xh+h^2+t)t+t^2]}{\sqrt{(2xh+h^2+t)^3}+\sqrt{t^3}} \\ &=\frac{(2x+0)[2x\cdot 0+0^2+t)^2+(2x\cdot 0+0^2+t)t+t^2]}{\sqrt{(2x\cdot 0+0^2+t)^3}+\sqrt{t^3}} \\ &=\frac{2x[3t^2]}{2\sqrt{t^3}} \\ &=3xt^{1/2} \\ &=3x\sqrt{x^2+y^2+z^2} \end{align} Note: At $*$ it was used: $$a^3-b^3=(a-b)(a^2+ab+b^2).$$