# How do you differentiate f(x)=sqrtcos(lnx^2) using the chain rule?

Nov 9, 2015

found: =1/x*(-sin(ln(x^2)))/sqrt(cos(ln(x^2))

#### Explanation:

Here, basically, you have 4 functions one inside the other! Starting for the most "external" you have:
the square root or ${\left(\right)}^{\frac{1}{2}}$, the $\cos$, then the $\ln$ and finally the square ${\left(\right)}^{2}$.
To derive you start from the outer one (the square root) and subsequently multiply by each derivative up the inner one (colored).
So:
given: $f \left(x\right) = {\left(\cos \left(\ln \left({x}^{2}\right)\right)\right)}^{\frac{1}{2}}$ you get:

$f ' \left(x\right) = \frac{1}{2} {\left(\cos \left(\ln \left({x}^{2}\right)\right)\right)}^{\frac{1}{2} - 1} \cdot \textcolor{red}{- \sin \left(\ln \left({x}^{2}\right)\right)} \cdot \textcolor{b l u e}{\frac{1}{x} ^ 2} \cdot \textcolor{g r e e n}{2 x} =$

rearranging:

$= {\left(\cos \left(\ln \left({x}^{2}\right)\right)\right)}^{- \frac{1}{2}} \cdot \left(- \sin \left(\ln \left({x}^{2}\right)\right)\right) \frac{1}{x} =$

=1/x*(-sin(ln(x^2)))/sqrt(cos(ln(x^2))