Find the derivative of $f(x)=7x\ln |x|$ Find the derivative of
$$f(x)=7x\ln|x|$$
How do they get the answer
$$f'(x)= 7\ln|x|+7$$
 A: Since $\dfrac{d}{dx}\vert x\vert=\dfrac{\vert x\vert}{x}$
\begin{eqnarray}
\frac{d}{dx}7x\ln\vert x\vert &=&7\ln\vert x\vert+7x\cdot\frac{1}{\vert x\vert}\cdot\frac{\vert x\vert}{x}\\
&=&7\ln\vert x\vert+7
\end{eqnarray}
Proof of derivative of $\vert x\vert$:
\begin{eqnarray}
\dfrac{d}{dx}\vert x\vert&=&\lim_{h\to0}\frac{\vert x+h\vert-\vert x\vert}{h}\\
&=& \lim_{h\to0}\frac{\vert x+h\vert-\vert x\vert}{h}\cdot\frac{\vert x+h\vert+\vert x\vert}{\vert x+h\vert+\vert x\vert}\\
&=&\lim_{h\to0}\frac{\vert x+h\vert^2-\vert x\vert^2}{h\cdot\left(\vert x+h\vert+\vert x\vert\right)}\\
&=&\lim_{h\to0}\frac{2xh+h^2}{h\cdot\left(\vert x+h\vert+\vert x\vert\right)}\\
&=&\lim_{h\to0}\frac{2x+h}{\vert x+h\vert+\vert x\vert}\\
&=&\frac{2x}{2\vert x\vert}\\
&=&\frac{x}{\vert x\vert}=\frac{\vert x\vert}{x}
\end{eqnarray}
A: You surely agree that the derivative of $\ln x$ (for $x>0$) is $1/x$.
The function $g(x)=\ln|x|$, defined for $x\ne0$, is even. Hence, by easy computations,
$$
g'(x)=-g'(-x)
$$
Indeed,
$$
\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=
\lim_{h\to0}-\frac{g(-x-h)-g(-x)}{-h}=-g'(-x)
$$
If $x<0$, then $-x>0$ and so $g'(-x)=\frac{1}{-x}$. Hence
$$
g'(x)=-g'(-x)=-\frac{1}{-x}=\frac{1}{x}
$$
Thus, for every $x\ne0$,
$$
g'(x)=\frac{1}{x}
$$
Thus, by the product rule, if $f(x)=7x\ln|x|$,
$$
f'(x)=7\left(\ln|x|+x\frac{1}{x}\right)=7(\ln|x|+1)
$$
