# Thread: implicit differentiation of an exponential function in two variables

1. ## implicit differentiation of an exponential function in two variables

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.

2. ## Re: implicit differentiation of an exponential function in two variables

Originally Posted by kingsolomonsgrave
$\displaystyle 2^{xy}(y+xy')\ln 2=y'+1$

3. ## 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?

4. ## Re: implicit differentiation of an exponential function in two variables

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 $\displaystyle y'=\dfrac{y2^{xy}}{1-x2^{xy}\ln 2}$ .