Implicit differentiation of $e^x-e^y-2^{xy}-1=0$

Question

I tried to do this:

$$e^x-e^y\frac{dy}{dx}...$$

I am stumped on '$-2^{xy}$', what it's deriative?

Answer 1

Hint: Here is the formula $$(a^u)'=u'a^u\ln a$$ then $$\dfrac{d}{dx}(2^{xy})=y2^{xy}\ln 2$$

Answer 2

$e^x-e^y-2^{xy}-1=0$ then write as $e^x-e^y-e^{xyln(2)}-1=0$ then differentiate

Answer 3

Use a combination of the chain rule an the product rule.

$u = xy\\ v = 2^u = 2^{xy}\\ \frac {dv}{dx} = (2^u)(\ln 2)\frac {du}{dx}\\ \frac {du}{dx} = y+ x\frac {dy}{dx}\\ \frac {dv}{dx} = (2^{xy})(\ln 2)(y + x\frac {dy}{dx})$

Answer 4

$$2^{xy}=e^{xy\ln 2}$$therefore $$e^x+e^y=e^{xy\ln 2}+1$$and by differentiating we obtain$$e^x+y'e^y=(y\ln 2+xy'\ln2)2^{xy}$$finally rearranging the terms leads to $$\LARGE y'=\dfrac{y\cdot2^{xy}\ln 2-e^x}{e^y-x\cdot2^{xy}\ln 2}$$