If $x=ye^y$, then explicit differentiation to find $\frac{dx}{dy}$, and the implicit differentiation to find $\frac{dy}{dx}$ yield consistent results.
Explain if the above line is True or False. (Ans: True)
But why? When we can clearly see using wolframalpha, explicit differentiation is $\dfrac{dx}{dy} = e^y \times (1 + y)$.
Implicit differentiation using wolframalpha, is $y'(x) = \dfrac{dy}{dx} = \dfrac{1}{e^y (y+1)}$
They seem clearly different, and not consistent result because they are reciprocal(?) of each other?
If $f(x)$ and $g(x)$ are differentiable functions and $y = f(x) \cdot g(x)$, the derivative of the product $y$ is given by:
$$\frac{dy}{dx} = \frac{d}{dx}\left(f(x) \cdot g(x)\right) = f'(x)g(x) + f(x)g'(x)$$
Thus, for $f(y) = y$ and $g(y) = e^y$, then the differentiation of $x= ye^y$ gives you:
$$\frac{dx}{dy} = \frac{d}{dy}\left(ye^y\right) = e^y + ye^y = e^y(1 + y)$$
Thus, $\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}} = \dfrac{1}{e^y(1 + y)}$.
Doing implicit differentiation on $x= ye^y$ with respect to $x$:
$$\frac{dx}{dx} = \frac{d}{dx}\left(ye^y\right) = \frac{dy}{dx}e^y + y\frac{(de^y)}{dx} = \frac{dy}{dx}e^y + ye^y\frac{dy}{dx} = e^y(1 + y)\frac{dy}{dx}$$ $$\therefore \ e^y(1 + y)\frac{dy}{dx} = \frac{dx}{dx} = 1 \ \Rightarrow \ \dfrac{dy}{dx} = \dfrac{1}{e^y(1 + y)}$$