Question

y = logx÷e^x differentiate it​

Answer 1

$y = log_{}(x) \div {e}^{x} \\ \\ \\y = \frac{ log_{}(x) }{ {e}^{x} }$

Now differentiating :-

$\frac{dy}{dx} = \frac{ {e}^{x} \frac{d}{dx}(logx) - logx \frac{d}{dx}( {e}^{x})}{ {({e}^{x})}^{2}} \\ \\ \\ \\ \\ \frac{dy}{dx} = \frac{ \frac{ {e}^{x} }{x} - logx {e}^{x} }{ {e}^{2x} } \\ \\ \\ \\ \frac{dy}{dx} = \frac{ {e}^{x}( \frac{1}{x} - logx) }{ {e}^{x} \times {e}^{x}} \\ \\ \\ \frac{dy}{dx} = \frac{(1 - xlogx)}{x {e}^{x} }$

The general Formula used :-

$y = {e}^{x} \\ \\ \frac{dy}{dx} = {e}^{x}$

$y = logx \\ \\ \\ \frac{dy}{dx} = \frac{1}{x}$

$y = \frac{u}{v} \\ \\ \\ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{ {v}^{2} }$

This is also a Function of a Function in division form.

Differentiation is done with respect to x when the value of y is given.

The functions are differentiated with certain general Formula.

But we can also solve them by classical mathematical way called as First Principle.

The direct way is easier and time saving.

The general expression of First principle is :-

$y=f(x)\\ \\ \\ \dfrac{dy}{dx} = Lt (h-->0) \dfrac{f(x+h)-f(x)} {h}$