# How to find differential of function of two variables : LOG[y,x] ?

• Feb 17th 2011, 06:34 AM
RCola
How to find differential of function of two variables : LOG[y,x] ?
I have function: z=(Log[y,x])^2 where y is base of log.
I need to find differential of function z regarding the x.
My solution:
[f(g(x))]'=f'(g(x))*g'(x)
so...
[(Log[y,x])^2]'=(2*Log[y,x])*(1/(x*Ln[y]))
but math program gives me following answer:
2Ln[x]/(x*Ln[y]^2)
What is the correct answer? Is there any mistake?
• Feb 17th 2011, 06:50 AM
SammyS
Quote:

Originally Posted by RCola
I have function: z=(Log[y,x])^2 where y is base of log.
I need to find differential of function z regarding the x.
My solution:
[f(g(x))]'=f'(g(x))*g'(x)
so...
[(Log[y,x])^2]'=(2*Log[y,x])*(1/(x*Ln[y]))
but math program gives me following answer:
2Ln[x]/(x*Ln[y]^2)
What is the correct answer? Is there any mistake?

Hi RCola.

If I understand correctly, you have $\displaystyle z=\left(\log_{\,y}x\right)^2$, and you need to find $\displaystyle \displaystyle {{\partial z}\over{\partial x}}\,.$

The change of base formula gives: $\displaystyle \displaystyle \log_{\,B} A = {{\ln A}\over{\ln B}}\ .$

$\displaystyle \displaystyle z=\left({{\ln x}\over{\ln y}}\right)^2$

BTW: I believe the answers are equivalent AND both are correct.
• Feb 17th 2011, 06:51 AM
e^(i*pi)
$\displaystyle z = (\log_y(x))^2 = \left(\dfrac{\ln(x)}{\ln(y)}\right)^2$
Treat $\displaystyle (\ln(y))^2$ as constant (because we're differentiating with respect to x)
$\displaystyle \dfrac{dz}{dx} = \dfrac{\frac{d}{dx} [\ln(x)]^2}{(\ln(y))^2}$