# How do you differentiate f(x) = sqrt(arctan(3x)  using the chain rule?

May 29, 2016

Chain rule says that $\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right] = f ' \left[g \left(x\right)\right] g ' \left(x\right)$.
Here we have three functions that are $3 x$, $\arctan$ and sqrt.
We start from the most "external" that is the square root
We know that

$\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2 \sqrt{x}}$

then, applying the chain rule we have

$\frac{d}{\mathrm{dx}} \sqrt{\arctan \left(3 x\right)} = \frac{1}{2 \sqrt{\arctan \left(3 x\right)}} \frac{d}{\mathrm{dx}} \arctan \left(3 x\right)$.

We have now to calculate the derivative of $\arctan \left(3 x\right)$

We know that

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

then we reapply the chain rule

$\frac{d}{\mathrm{dx}} \arctan \left(3 x\right) = \frac{1}{1 + {\left(3 x\right)}^{2}} \frac{d}{\mathrm{dx}} 3 x$

and finally the easy one

$\frac{d}{\mathrm{dx}} 3 x = 3$.

We substitute everything back and write

d/dxsqrt(arctan(3x))=3/(2sqrt(arctan(3x))(1+9x^2)