Question

(X^2+2x-3) logx derivative

Answer 1

Question :-

Find the derivative of -

$\to \sf \: \: ( {x}^{2} + 2x - 3) \log x$

Answer :-

$\to \boxed{ \sf \frac{dy}{dx} = \log x(2x + 2) + x + 2 - \frac{3}{x} } \:$

Used Concept :-

Here we will use product rule of derivative i.e

$\to \sf \frac{d}{dx} \: u \: v = u \: \frac{dv}{dx} + v \: \frac{du}{dx}$

Here , u and v are function of x .

Solution :-

We have and let ,

$\to \sf \: y = ( {x}^{2} + 2x - 3) \log x \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \: \\ \to \sf \frac{dy}{dx} = ( \log x) \frac{d}{dx} ( {x}^{2} + 2x - 3) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf +( {x}^{2} + 2x - 3) \frac{d}{dx} \log x \\ \\ \boxed{ \because \sf \: \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \: \: \: } \\ \boxed{\sf \because \: \frac{d}{dx} \log x = \frac{1}{x} \: \: \: \: \: } \\ \boxed{\sf\because \frac{d}{dx} constant = 0 } \\ \\ \to \sf \frac{dy}{dx} = \log x(2x + 2) + \frac{ {x}^{2} + 2x - 3}{x} \\ \bf or \\ \\ \to \boxed{ \sf \frac{dy}{dx} = \log x(2x + 2) + x + 2 - \frac{3}{x} }$