Hi he told us we can split $\displaystyle cos^2 $ into $\displaystyle cos . cos$ and use the chain rule. But i'm still not getting the same answer as him. Thanks
Why don't you make some effort to learn LaTeX?
It is so hard to read you posts.
If $\displaystyle y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} $ then
$\displaystyle \dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} }}$
$\displaystyle u=d^2+l^2+2dlcos(\phi)cos(\phi)$
I reckon the reason he said to split $\displaystyle cos^2(\phi)$ into $\displaystyle cos(\phi)*cos(\phi)$
is to conveniently apply the product rule to differentiating $\displaystyle cos^2(\phi)$ while you are getting used to the chain rule.
The chain rule is used in two ways along with the product rule in that case.
If you did not want to memorise the rule given by Plato then using the chain rule
$\displaystyle \displaystyle y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} = (d^2 + l^2 + dl\cos ^2 (\phi ))^{\frac{1}{2}}$
let $\displaystyle \displaystyle u = d^2 + l^2 + dl\cos ^2 (\phi ) \implies y = u^{\frac{1}{2}}$
Finally $\displaystyle \displaystyle \frac{dy}{d\phi} = \frac{du}{d\phi}\times \frac{dy}{du}$
Following pickslides post above,
then you can use the product rule on $\displaystyle \displaystyle\frac{du}{d\phi}$
by writing $\displaystyle cos^2\phi=cos\phi*cos\phi$
though you don't need to.
You could also apply the chain rule to that.
Better again, learn Plato's method.
One thing that's bothering adam_leeds is that Plato has a minor typo in his expressions, but they are consistent with each other. (Plato did mention that using LaTeX would make the problem more readable.)
They should read:
"If $\displaystyle y = \sqrt {d^2 + l^2 + 2dl\cos ^2 (\phi )} $ then
$\displaystyle \dfrac{{dy}}{{d\phi }} = \dfrac{{4dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + 2dl\cos ^2 (\phi )} }}$"