Exponential and Logarithmic Differentiation.

Question

Q. If $xe^{xy}=y+\sin^2x$, then find $\frac{dy}{dx}$ at x=0.

If we differentiate the function directly as follows:

$e^{xy}+xe^{xy}\left[y+x\frac{dy}{dx}\right]=\frac{dy}{dx}+\sin\left(2x\right)$

At x=0 We have

$1+0=\frac{dy}{dx}+0$ which gives

$\frac{dy}{dx}=1$

But if we apply logarithm and differentiate we have:

$\log\left(x\right)+xy=\log\left(y+\sin^2x\right)$ which when differentiated gives

$\frac{1}{x}+y+x\frac{dy}{dx}=\frac{\frac{dy}{dx}+\sin2x}{y+\sin^2x}$

If we take x=0 here, we get undefined values. And we are unable to obtain the value of dy/dx individually. Is there any mistake in the logarithmic method?

Answer 1

If x=0 then log(x) is undefined. If you continue and solve for dy/dx you get the indeterminate form 0/0 as x approaches 0.

Answer 2

if you at the constraint $$xe^{xy} = y + \sin^2 x $$ you see that at $x = 0,$ gives $y = 0.$ now for $$ y = x + \cdots \text{ for } x = 0+\cdots$$ this implies $\dfrac{dy}{dx} = 1$ at $x = , y = 0.$

$\bf edit:$

even you logarithmic differentiation result gives you the same. here is the reason why. you have $$\frac1x + y + x\frac{dy}{dx} = \dfrac{\frac{dy}{dx}+\sin 2x}{y+\sin^2x} $$ if you look at the behavior off $y$ for $x = 0 + \cdots,$ we find that $$\frac1x \cdots= \frac{1}{y} +\cdots$$ again $y = x + \cdots$ balances the dominant term $\frac1x$ if we take $\frac{dy}{dx} = 1+\cdots$