# differentiating $\frac{5x}{1 + x^2}$ using the power rule

I am trying to differentiate $s=\frac{5x}{1+x^2}$. I have decided to tackle this by using the power rule if $y=x^n$ $$\dfrac {dy}{dx}=nx^{n-1}$$

This gives me $$\dfrac {dy}{dx}=\frac{5}{2x}$$

I got rid of the 1 in the denominator because $0 \times 1^{1-1}$=$0$

Given the answer in the textbook, I assume this is incorrect. Can anybody tell me why?

• It's not just a variable to a power...it's a rational function; $f(x)/g(x)$. You need to use the quotient rule... – Iceman Dec 20 '15 at 12:01

## 3 Answers

Can anybody tell me why?

Because in general $$\frac{d}{dx} \left(\frac{g(x)}{h(x)}\right)\neq \frac{\frac{d}{dx}g(x)}{\frac{d}{dx}h(x)}$$ (From your post, it seems you just calculated the derivative of numerator and the derivative of denominator.)

In this case, you have to combine the power rule with other rules (e.g. product rule and chain rule): \begin{align}\frac{d}{dx}\left(\frac{5x}{1+x^2}\right)&=\frac{d}{dx}\left(5x(1+x^2)^{-1}\right)\\\\ &=\left(\frac{d}{dx}(5x)\right)(1+x^2)^{-1}+5x\frac{d}{dx}((1+x^2)^{-1})\\\\ &=5(1+x^2)^{-1}+5x(-1)(1+x^2)^{-2}(2x)\\\\ &= -\frac{5 (x^2-1)}{(x^2+1)^2} \end{align}

the Quotient rule is applicable here since you have a rational function. The quotient rule is, for function $h(x)=f(x)/g(x)$

$$h'(x)=\frac{g(x)\cdot f'(x)-f(x)\cdot g`(x)}{[g(x)]^2}$$

take $f(x)=5x, g(x)=1+x^2$.

So, we have

$$h'(x)=\frac{(1+x^2)\cdot 5-5x\cdot 2x}{(1+x^2)^2}$$

which you can now reduce...

You have to use product rule which says $$\frac{d}{dx}(f(x)g(x)) = (f(x)g(x))' = (fg)' =f'g+fg' = f'(x)g(x) + f(x)g'(x)$$ and also composite function derivative which says $$(g(h(x)))' = f'(h(x))\cdot h'(x)$$

You got $f(x)=5x$, $g(x) = x^{-1}$ and $h(x)=1+x^2$, so $$\frac{d}{dx}(f(x)g(h(x))) = f'(x)g(h(x)) + f(x)(g'(h(x))h'(x)$$

Thus $$\frac{d}{dx}\frac{5x}{1+x^2} = 5x (1+x^2)^{-1} = (5x)'(1+x^2)^{-1} + 5x((1+x^2)^{-1})' = \frac{5}{1+x^2} - \frac{5x\cdot 2x}{(1+x^2)^2}$$