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implicit differentiation of an exponential function in two variables

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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.

Re: implicit differentiation of an exponential function in two variables

Quote:

Originally Posted by

**kingsolomonsgrave**

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

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?

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