# Parital derivation

Discussion in 'Math' started by mo2015mo, May 11, 2013.

1. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
Hi guys ,

I have been having problems with the attached question, I have tried to solve it But don't get the correct answer
i want to know HOW can derive the function under the root square
plz help meee

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2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
How come this isn't in homework?

Have you heard of implicit differentiation?

Square both sides, for example

${y^2} = uv$

Differentiate with respect to u, holding v constant.

$\begin{array}{l}
2y\frac{{\partial y}}{{\partial u}} = \frac{{\partial (uv)}}{{\partial u}} = v \\
\frac{{\partial y}}{{\partial u}} = \frac{v}{{2y}} = \frac{v}{{2\sqrt {uv} }} = \frac{{\sqrt v }}{{2\sqrt u }} \\
\end{array}$

If you are not familiar with implicit differentiaion, you can do this one the ordinary way as follows

$y = \sqrt {uv} = {u^{\frac{1}{2}}}{v^{\frac{1}{2}}}$

Differentiate with respect to u, holding v constant.

$\begin{array}{l}
y = \sqrt {uv} = {u^{\frac{1}{2}}}{v^{\frac{1}{2}}} \\
\frac{{\partial y}}{{\partial u}} = \frac{1}{2}{u^{ - \frac{1}{2}}}{v^{\frac{1}{2}}} = \frac{{\sqrt v }}{{2\sqrt u }} \\
\end{array}$

As before.

In this instance implicit differentiation offers no real advantages, but sometimes it is the only way and often it is the easiest way with a square root involved.

Does this help?

Last edited: May 12, 2013
mo2015mo likes this.
3. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
yes yes i have heard of implicit differentiation ,But it did not come into My mind
Studiot
Thank you so much for all youve done to help me

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Presumably you can finish all the rest off now?

5. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
yes, i have solved the rest and gotten the correct answer

Thanks again for help me

6. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Good. I would recommend doing them directly, as well, and verifying that you get the same answers.