# Math Help - Simple differentiation not getting lecturers answer :(

1. ## Simple differentiation not getting lecturers answer :(

Hi he told us we can split $cos^2$ into $cos . cos$ and use the chain rule. But i'm still not getting the same answer as him. Thanks

2. Why don't you make some effort to learn LaTeX?
It is so hard to read you posts.

If $y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )}$ then

$\dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} }}$

Hi he told us we can split $cos^2$ into $cos . cos$ and use the chain rule. But i'm still not getting the same answer as him. Thanks
It's difficult to explain how to get the lecturer's answer if you don't give us that answer.

4. Originally Posted by Plato
Why don't you make some effort to learn LaTeX?
It is so hard to read you posts.

If $y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )}$ then

$\dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} }}$
Sorry, thanks.

My lecturer got it to be negative and the two on the denominator was at the front of $dlcos^2(\phi)$

You have missed off the two in the question. But i'll try and do yours then do it to mine and see if i get the same answer.

5. Originally Posted by Plato
Why don't you make some effort to learn LaTeX?
It is so hard to read you posts.

If $y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )}$ then

$\dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} }}$
I don't really understand how you do the chain rule bit

is u = $(d^2 + l^2 + dlcos(\phi))^(1/2))$ and v = $(cos(\phi))^(1/2)$

I don't really understand how you do the chain rule bit

is u = $(d^2 + l^2 + dlcos(\phi))^(1/2))$ and v = $(cos(\phi))^(1/2)$
$u=d^2+l^2+2dlcos(\phi)cos(\phi)$

I reckon the reason he said to split $cos^2(\phi)$ into $cos(\phi)*cos(\phi)$

is to conveniently apply the product rule to differentiating $cos^2(\phi)$ 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.

7. Originally Posted by Archie Meade
$u=d^2+l^2+2dlcos(\phi)cos(\phi)$
How do you differentiate that with the product rule when it is all square rooted?

8. Originally Posted by Plato
Why don't you make some effort to learn LaTeX?
It is so hard to read you posts.

If $y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )}$ then

$\dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} }}$
Please explain how you did this

Please explain how you did this
First I did not use u-substitution. I learned basic formulas as I went along.
It appears that you may have skipped that basic learning process.

$\left( {\sqrt {f(x)} } \right)^\prime = \dfrac{{f'(x)}}
{{2\sqrt {f(x)} }}$

10. Originally Posted by Plato
First I did not use u-substitution. I learned basic formulas as I went along.
It appears that you may have skipped that basic learning process.

$\left( {\sqrt {f(x)} } \right)^\prime = \dfrac{{f'(x)}}
{{2\sqrt {f(x)} }}$
Yep done it now, thank you.

Please explain how you did this
If you did not want to memorise the rule given by Plato then using the chain rule

$\displaystyle y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} = (d^2 + l^2 + dl\cos ^2 (\phi ))^{\frac{1}{2}}$

let $\displaystyle u = d^2 + l^2 + dl\cos ^2 (\phi ) \implies y = u^{\frac{1}{2}}$

Finally $\displaystyle \frac{dy}{d\phi} = \frac{du}{d\phi}\times \frac{dy}{du}$

How do you differentiate that with the product rule when it is all square rooted?
Following pickslides post above,

then you can use the product rule on $\displaystyle\frac{du}{d\phi}$

by writing $cos^2\phi=cos\phi*cos\phi$

though you don't need to.
You could also apply the chain rule to that.

Better again, learn Plato's method.

13. Originally Posted by Plato
Why don't you make some effort to learn LaTeX?
It is so hard to read you posts.

If $y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )}$ then

$\dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + dl\cos ^2 (\phi )} }}$

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.)

"If $y = \sqrt {d^2 + l^2 + 2dl\cos ^2 (\phi )}$ then
$\dfrac{{dy}}{{d\phi }} = \dfrac{{4dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2 + l^2 + 2dl\cos ^2 (\phi )} }}$"