A partial differentiation

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February 20th 2010, 01:28 PM
Lex

A partial differentiation

Hi,

I am trying to differentiate
(N*k/3)*[T-(Q/6.7)*arctan(6.7T/Q)]
with respect to T.

Apparently the answer is
(N*k/3)*[(6.7T/Q)^2 / 1 + (6.7T/Q)^2]

I am getting
(N*k/3)*[1/(1+(6.7T/Q)^2)]

Could someone give me breakdown of how this is done and perhaps point out where I have gone wrong?

Thanks

(P.S. sorry for not being able to use MATH)
February 20th 2010, 02:00 PM
o_O

Is this it: $F = \frac{Nk}{3}\left[ {\color{red}T} - \frac{Q}{6.7}\arctan \left(\frac{6.7{\color{red}T}}{Q}\right)\right]$ ?

It's kind of hard to tell from what you've written. Anyway, assuming that it is, to differentiate with respect to $T$ , treat every other variable as constants.

So: $\frac{\partial F}{\partial T} = \frac{Nk}{3} \left[ {\color{red}1} - \frac{Q}{6.7}\underbrace{\left( \frac{\frac{6.7}{Q}}{1+ \left(\frac{6.7T}{Q}\right)^2}\right) }_{\displaystyle \text{Since: } \frac{d}{dx} \arctan ({\color{red}u}) = \frac{{\color{red}u}'}{1+{\color{red}u}^2}}\right]$
February 20th 2010, 02:10 PM
Archie Meade

Quote:

Originally Posted by Lex

Hi,

I am trying to differentiate
(N*k/3)*[T-(Q/6.7)*arctan(6.7T/Q)]
with respect to T.

Apparently the answer is
(N*k/3)*[(6.7T/Q)^2 / 1 + (6.7T/Q)^2]

I am getting
(N*k/3)*[1/(1+(6.7T/Q)^2)]

Could someone give me breakdown of how this is done and perhaps point out where I have gone wrong?

Thanks

(P.S. sorry for not being able to use MATH)

Hi Lex,

$\left(N\frac{k}{3}\right)\frac{d}{dT}\left[T-\left(\frac{Q}{6.7}\right)Tan^{-1}\left(\frac{6.7T}{Q}\right)\right]$

$\frac{d}{dx}Tan^{-1}\frac{x}{a}=\frac{a}{a^2+x^2}$

$a=\frac{Q}{6.7},\ x=T$

The derivative is

$\left(N\frac{k}{3}\right)\left[1-a\frac{a}{x^2+a^2}\right]$ $=\left(N\frac{k}{3}\right)\left[\frac{x^2+a^2-a^2}{x^2+a^2}\right]$

$=\left(N\frac{k}{3}\right)\left[\frac{T^2}{T^2+\left(\frac{Q}{6.7}\right)^2}\right]=\left(N\frac{k}{3}\right)\left[\frac{\left(\frac{6.7T}{Q}\right)^2}{1+\left(\frac {6.7T}{Q}\right)^2}\right]$
February 21st 2010, 05:51 AM
Lex

Thanks to all.

It appears that my calculus was correct, but it was the rearrangement that eluded me.