How do I differentiate $3x/(x^2+1)$?

Question

I kind of understand how the power rule works. You multiply N and subtract 1 from N in the exponent. How do I differentiate $$y=\frac{3x}{x^2+1}$$

Answer 1

To differentiate function which is fraction of two functions of $x$, i.e; of the form $y=u(x)/v(x)$, then, $$y'=\frac{v(x)\cdot\frac{du(x)}{dx}-u(x)\cdot\frac{dv(x)}{dx}}{v(x)^2}$$

See here for more detailed proof.

Here, you have $u(x)=3x$ and $v(x)=x^2+1$, as, you know how to differentiate exponents of $x$, $\frac{dv(x)}{dx}=2x$. Just plug in the values.

Answer 2

You use the Quotient Rule. $$\biggr(\frac{f}{g}\biggr)’ = \frac{f’g-g’f}{g^2} \text{ if } g^2 \neq 0$$ Here, $f$ means $f(x)$ and $g$ means $g(x)$, so simply, they are two functions.

Your function is $$y = \frac{3x}{x^2+1}.$$ You can solve it by breaking it down into $f(x)$ and $g(x)$ and then using the Quotient Rule.

$$\text{Say } f(x) = 3x \text{ and } g(x) = x^2+1 \text{ so that } y = \frac{f(x)}{g(x)}.$$

$$y’ = \biggr(\frac{3x}{x^2+1}\biggr)’ = \frac{(3x)’\cdot(x^2+1)-(x^2+1)’\cdot3x}{(x^2+1)^2}$$

$$\frac{d}{dx} 3x = 3 \text{ and } \frac{d}{dx} (x^2+1) = 2x$$

Simplify.

$$y’ = \frac{3\cdot(x^2+1)-2x\cdot(3x)}{(x^2+1)^2} \implies y’ = \frac{3x^2+3-6x^2}{(x^2+1)^2} \implies \boxed{y’ = \frac{-3(x^2-1)}{(x^2+1)^2}}$$

There isn’t really any good simplification from here on, so just leave the answer in factored form.

Answer 3

you have the function: $$y=\frac{3x}{x^2+1}$$ you really have two options, to use the "product rule + chain rule" or to use the combined "quotient rule". I will use the quotient rule which states for a function: $$y=\frac{u}{v}$$ $$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$ so if: $$y=\frac{3x}{x^2+1}$$ then: $$\frac{dy}{dx}=\frac{(x^2+1)(3)-(3x)(2x)}{(x^2+1)^2}=\frac{3-3x^2}{(x^2+1)^2}=\frac{3(1-x^2)}{(1+x^2)^2}$$

Lets now try and take a different approach:

as I mentioned earlier there are two other rules, the chain rule: $$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$$ and the product rule: $$y=f(x)g(x)$$ $$\frac{dy}{dx}=f'(x)g(x)+f(x)g'(x)$$ we can then view our function instead as: $$y=(3x)(x^2+1)^{-1}$$ and use a combination of these two rules