# How do you differentiate (3x^2+4) / (sqrt(1+x^2))?

Apr 5, 2015

Hey there! :)

Looks tough when you first see it, right? However, differentiation rules are like super powers.

Here is the short story:

$\frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2} + 4}{\sqrt{1 + {x}^{2}}}\right) = \frac{6 x}{\sqrt{1 + {x}^{2}}} - \frac{x \left(3 {x}^{2} + 4\right)}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

Now for the far more exciting long story:

The quotient rule says that if we have two differentiable functions, say $f \left(x\right) \mathmr{and} g \left(x\right)$ such that $g \left(x\right) \ne 0$, then we have

$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) \cdot g \left(x\right) - f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}{g \left(x\right)} ^ 2$

So now we just apply this!

So,

$\frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2} + 4}{\sqrt{1 + {x}^{2}}}\right) = \frac{\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 4\right) \cdot \sqrt{1 + {x}^{2}} - \left(3 {x}^{2} + 4\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{1 + {x}^{2}}\right)}{\sqrt{1 + {x}^{2}}} ^ 2$

Now we have to use a few super powers here. The power rule and chain rule are part of our arsenal.

I hope it is clear that

$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 4\right) = 2 \cdot 3 x + 0 = 6 x$

(If not, just shout!) - We just applied the power rule to the first term and the second term is a constant. And constants don't stand a chance against differentiation! They just become zero.

Next, we harness the power of the chain rule, that says for any differentiable functions, say, $u \left(x\right)$ and $h \left(x\right)$,

$\frac{d}{\mathrm{dx}} \left(u \left(h \left(x\right)\right)\right) = \frac{d}{\mathrm{dh} \left(x\right)} \left(u \left(h \left(x\right)\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(h \left(x\right)\right)$

We apply it (here u is the square root function and h(x) = (1+x^2)):

$\frac{d}{\mathrm{dx}} \left(\sqrt{1 + {x}^{2}}\right) = \frac{1}{2} \cdot {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(2 x\right) = x {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}$

Finally, plugging these things back in their rightful place, we get

$\frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2} + 4}{\sqrt{1 + {x}^{2}}}\right) = \frac{6 x \cdot \sqrt{1 + {x}^{2}} - \left(3 {x}^{2} + 4\right) \cdot x {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}}{1 + {x}^{2}}$
$= \frac{6 x \sqrt{1 + {x}^{2}}}{1 + {x}^{2}} - \frac{x \left(3 {x}^{2} + 4\right)}{\left(1 + {x}^{2}\right) \cdot \sqrt{1 + {x}^{2}}}$
$= \frac{6 x}{\sqrt{1 + {x}^{2}}} - \frac{x \left(3 {x}^{2} + 4\right)}{{\left(1 + {x}^{2}\right)}^{\frac{3}{2}}}$