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I have a practice problem that says:

Graph

Let $$j(x) = \frac{g(x)}{f(x)}$$

Find $j'(1)$:

I don't know how to do this. The answer the book has is $-2$. What I tried to do was set $j(0) = \frac{g(0)}{f(0)}= 0$ because the derivative of a constant (such as $1$) is $0$.

Can someone explain what I have to do?

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  • $\begingroup$ what else is given? $\endgroup$ – Dr. Sonnhard Graubner Apr 8 '15 at 14:23
  • $\begingroup$ Quotient rule? $\space$ $\endgroup$ – graydad Apr 8 '15 at 14:24
  • $\begingroup$ I forgot the graph, I apologize. I'll upload it in a few seconds. $\endgroup$ – 2redgoose Apr 8 '15 at 14:24
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    $\begingroup$ The answer depends on $f$ and $g$. You want to know the rule for $(g/f)'$. $\endgroup$ – mvw Apr 8 '15 at 14:25
  • $\begingroup$ Also, $j(0) = 0$ doesn't imply $j'(0) = 0$, since $j$ is not constant. $\endgroup$ – Paul Picard Apr 8 '15 at 14:29
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$$j'(x)=\frac{f(x)g'(x)-g(x)f'(x)}{f^{2}(x)}$$ So $$j'(1)=\frac{f(1)g'(1)-g(1)f'(1)}{f^{2}(1)}=\frac{2\times (-1)-3\times 2}{4}=-2$$

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Given the graph, we can write that for $0 \lt x \leq 2$, we have : $$f(x) = 2x$$$$g(x) = 4-x$$ Hence (since we exclude $x=0$) : $$j(x) = \frac{4-x}{2x} = \frac{2}{x} - \frac{1}{2}$$ Now, for $0 \lt x \leq 2$, $j$ is continuous, so you can derivate the expression and you have : $$j'(x) = -\frac{2}{x^2}$$

And finally, $$j'(1) = -\frac{2}{1} = -2$$

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