# implicit differentiation of an exponential function in two variables

• Oct 23rd 2012, 10:15 PM
kingsolomonsgrave
implicit differentiation of an exponential function in two variables
Attachment 25383

for the RHS how would I tackle this?

I know for taking the derivative of $\displaystyle a^x$ would give $\displaystyle a^x*ln(a)$ but im not sure how to apply that in this situation, given that one of the exponential variables (y) is itself a function of x.
• Oct 23rd 2012, 10:22 PM
FernandoRevilla
Re: implicit differentiation of an exponential function in two variables
Quote:

Originally Posted by kingsolomonsgrave

$\displaystyle 2^{xy}(y+xy')\ln 2=y'+1$
• Oct 24th 2012, 07:25 AM
kingsolomonsgrave
Re: implicit differentiation of an exponential function in two variables
thanks!

how would I solve for y' in this case? Just multipy by 2^xy, group like terms and solve as normal?
• Oct 25th 2012, 12:47 AM
FernandoRevilla
Re: implicit differentiation of an exponential function in two variables
Quote:

Originally Posted by kingsolomonsgrave
how would I solve for y' in this case? Just multipy by 2^xy, group like terms and solve as normal?

Solve it as a first degree equation on the unknown $\displaystyle y'$. You'll obtain $\displaystyle y'=\dfrac{y2^{xy}}{1-x2^{xy}\ln 2}$ .