# partial derivative question

• Apr 25th 2010, 09:16 PM
dben
partial derivative question
seems pretty simple but the $\displaystyle tan(\phi)$ on the left side threw me off

$\displaystyle tan(\phi) = \frac {X_L}{R}$ ; find $\displaystyle \frac {\partial \phi}{\partial R}$

Answer key states in the first step that:

$\displaystyle \frac {\partial \phi}{\partial R}= \frac {1}{(\frac {X_L}{R^2}) + 1} * \frac {-X_L}{R^2}$

not sure how they got $\displaystyle \frac {1}{(\frac {X_L}{R^2}) + 1}$
• Apr 25th 2010, 09:26 PM
Random Variable
EDIT: $\displaystyle \sec^{2} \phi \ \frac{d \phi}{d R} = - \frac{X_{L}}{R^{2}}$

but $\displaystyle \sec^{2} \phi = 1+\tan^{2}\phi$

so $\displaystyle \frac{d \phi}{d R} = - \frac{1}{1+\tan^{2} \phi}\frac{X_{L}}{R^{2}} = - \frac{1}{1+(\frac{X_{L}}{R})^{2}} \frac{X_{L}}{R^{2}}$
• Apr 25th 2010, 09:58 PM
dben
edit: when you substitute $\displaystyle \tan^2 \phi$ for $\displaystyle \frac {X_L}{R}$ the whole fraction becomes squared, however, in the answer just the $\displaystyle R$ in the denominator is squared, not the $\displaystyle X_L$
• Apr 26th 2010, 09:58 PM
Random Variable
Quote:

Originally Posted by dben
edit: when you substitute $\displaystyle \tan^2 \phi$ for $\displaystyle \frac {X_L}{R}$ the whole fraction becomes squared, however, in the answer just the $\displaystyle R$ in the denominator is squared, not the $\displaystyle X_L$

It has to be a misprint.
• Apr 26th 2010, 10:37 PM
dben
yea, i put it down for the day and looked at it again tonight and realized it is a misprint. The final answer that is printed only works if the whole fraction is squared...Thanks!

dave