Finding derivative of three variables, determining which of two answers is correct

Question

I'm trying to find $\dfrac{\partial w}{\partial x}$ for $$w = \ln\left(\sqrt{x^2+y^2+z^2}\right)$$

As $\dfrac{d\ln(x)}{dx} = \dfrac{x'}{x}$, I get $$\frac{1}{\sqrt{x^2+y^2+z^2}}\cdot x$$ as the derivative of the square root part.

So $$\frac{\frac{x}{\sqrt{x^2+y^2+z^2}}}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\sqrt{x^2+y^2+z^2}}\cdot\frac{\sqrt{x^2+y^2+z^2}}{1}= x$$

I put it into the symbolab calculator to check and it showed $$\frac{x}{\sqrt{x^2+y^2+z^2}}\cdot\frac{1}{\sqrt{x^2+y^2+z^2}}$$

Which doesn't seem right. Now I'm wondering if I'm missing something or it's wrong. Can someone confirm which is correct?

Answer 1

You error lies between the third to last step and the second to last step.$\frac{\frac a b}c=\frac a {bc}$.

Answer 2

I suppose that life would have been easier writing before anything else $$w = \ln\left(\sqrt{x^2+y^2+z^2}\right)=\frac 12 \ln\left({x^2+y^2+z^2}\right)$$ from which you deduce that $$\dfrac{\partial w}{\partial x}=\frac 12 \frac{2x}{x^2+y^2+z^2}=\frac{x}{x^2+y^2+z^2}$$