0
$\begingroup$

Hey StackExchange I have a question for you guys. I have a homework problem and my first intuition is to use the quotient rule (or re-write the expression to use the product rule) but the product/quotient rules haven't been covered yet so I feel like they wouldn't expect me to use them. Perhaps you guys could show me where to start. The problem is as follows:

If $F(X) = \frac{5x}{1 + x^2}$, find $F'(2)$ and use it to find an equation of the tangent line to the curve $y = \frac{5x}{1+x^2}$ at the point $(2, 2)$.

I know I just have to calculate the derivative and plug in numbers, my question just concerns alternatives to the quotient rule.

$\endgroup$
1
  • $\begingroup$ They probably want you to do it using the definition of the derivative using limits. $\endgroup$
    – Mastrel
    Jul 7 '14 at 3:09
1
$\begingroup$

I suspect that they intend for you to find $$\lim_{x\to 2}\frac{F(x)-F(2)}{x-2},$$ which is the limit definition of $F'(2).$

$\endgroup$
4
  • $\begingroup$ I hope you can help me out, I'm stuck. Perhaps I'm not following the right steps. I set the problem up as follows: $$\frac{\frac{5x}{1+x^2} - \frac{5a}{1+a^2}}{x-a}$$ It becomes: $$\frac{5x(1+a^2)-[5a(1+x^2)]}{(1+x)^2(1+a)^2(x-a)}$$ Which turns into: $$\frac{5x+5xa^2-[5a+5ax^2]}{(1+x)^2(1+a)^2(x-a)}$$ And thus: $$\frac{5(x+xa^2)-[5(a+ax^2)]}{(1+x)^2(1+a^2)(x-a)}$$ And now I'm stuck. I feel like I'm not following the right path, any suggestions? $\endgroup$ Jul 7 '14 at 3:44
  • $\begingroup$ You're almost there in your second-to-last step! Instead, rearrange it as $$\frac{5x-5a+5a^2x-5ax^2}{x-a},$$ and try factoring the top by grouping from there. See if that gets you the rest of the way, but let me know if you get stuck again. $\endgroup$ Jul 7 '14 at 3:48
  • $\begingroup$ But what about my other denominator factors? Where did they go? $\endgroup$ Jul 7 '14 at 3:55
  • $\begingroup$ Oops! That's what I get for trying to do this while half-asleep. That should be $$\frac{5x-5a+5a^2x-5ax^2}{(1+x^2)(1+a^2)(x-a)},$$ instead. $\endgroup$ Jul 7 '14 at 10:50
1
$\begingroup$

As an alternative to the quotient rule, you can always try logarithms.

$$y=\frac{5x}{1+x^2}\\ \ln (y)=\ln (5x)-\ln(1+x^2)\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}-\frac{2x}{1+x^2} $$

and now multiply by $y$ and substitute in your values of $x$ and $y$.

It works out the same as using the quotient rule, since you can always derive the quotient rule by using logs in this way.

$\endgroup$
0
$\begingroup$

Knowing that $F(x)=5x/(1+x^{2})$ is differentiable, $$ F(x+\Delta x)=F(x)+F'(x)\Delta x + o(\Delta x) = \frac{5x+5\Delta x}{1+(x+\Delta x)^{2}}, $$ where $o(\Delta x)$ means terms that vanish faster than $\Delta x$ as $\Delta x\rightarrow 0$. Clearing denominators: $$ (F(x)+F'(x)\Delta x+o(\Delta))\cdot(1+x^{2}+2x\Delta x+o(\Delta x))=5x+5\Delta x . $$ Collecting $\Delta x$ terms gives $$ F(x)2x+F'(x)(1+x^{2})=5 \\ \implies F'(x)=\frac{5-2xF(x)}{1+x^{2}}. $$ You want $F'(2)$ which involves $F(2)=10/5=2$ and is $F'(2)=(5-2(2)(2))/5=-3/5$.

$\endgroup$
0
$\begingroup$

The quotient rule is just a shortcut for the product rule and chain-rule. $$\frac {f(x)}{g(x)} = f(x) \cdot \underbrace{g(x)^{-1}}_{\text{Polynomial}}$$

So $$\begin{align}\frac{d\frac{f(x)}{g(x)}}{dx} &= \frac{d}{dx}\left(f(x)\cdot g(x)^{-1}\right) \\ &\overbrace{=}^{\text{product rule}} f'(x)g(x)^{-1} + f(x)\frac{d}{dx} g(x)^{-1} \\ & \overbrace{=}^{\text{chain rule}} f'(x)g(x)^{-1} + f(x)\cdot -1 \cdot g(x)^{-2} \cdot g'(x) \\ &= \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\end{align}$$

So if you want you can "just" use product and chain rule. Anyway, just because your book hasn't covered something doesn't mean it isn't true yet. Since you know the chain rule, you may as well go ahead and use it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.