Parital derivation

Thread Starter

mo2015mo

Joined May 9, 2013
157
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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studiot

Joined Nov 9, 2007
4,998
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:

Thread Starter

mo2015mo

Joined May 9, 2013
157
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 :)
 
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