Find the derivative of $$f(x)=7x\ln|x|$$ How do they get the answer $$f'(x)= 7\ln|x|+7$$
Find the derivative of $f(x)=7x\ln |x|$
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derivatives
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2Do you know the product rule? – 2017-01-31
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4Isn't it supposed to be $7\ln |x|+7$? – 2017-01-31
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0Arthur - I think you're right – 2017-01-31
2 Answers
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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}
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0How do you get the part + 7x * 1/ IxI * IxI/x ? – 2017-01-31
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0From the chain rule. $[f(g(x))]^\prime=f^\prime(g(x))\cdot g^\prime(x)$. – 2017-01-31
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0You are welcome. – 2017-02-01
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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 $f(x)=7x\ln|x|$, $$ f'(x)=7\left(\ln|x|+x\frac{1}{x}\right)=7(\ln|x|+1) $$