Question

find the derivative of :-​

Answer 1

Answer:

See the above solution.

Answer 2

$\sf \rightarrow \quad y = cot \bigg(\pi - \dfrac{1}{x} \bigg) \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \frac{d \: cot \big(\pi - \frac{1}{x} \big)}{dx} \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \frac{d \: cot \big(\pi - \frac{1}{x} \big)}{d \bigg( \pi - \dfrac{1}{x} \bigg) } \cdot \frac{d \bigg( \pi - \dfrac{1}{x} \bigg) }{dx} \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = -{cosec}^{2} \bigg( \pi - \dfrac{1}{x} \bigg) \cdot \bigg(\frac{d\pi}{dx} - \frac{d {x}^{ - 1} }{dx} \bigg) \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = -{cosec}^{2} \bigg( \pi - \dfrac{1}{x} \bigg) \cdot \{0 - ( - {x}^{ - 2} ) \} \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \frac{-{cosec}^{2} \big( \pi - \frac{1}{x} \big)}{ {x}^{2} }$

$\sf \underline{Formulae \: Used} \\ \star \qquad \frac{du}{dt} = \frac{du}{dx} \cdot \frac{dx}{dt} \\ \\ \star \qquad \frac{da}{dt} = 0 \qquad a = \sf constant \\ \\ \star \qquad \frac{d \: cot \: x}{dx} = - {cosec}^{2} \: x \\ \\ \star \qquad \frac{d(u - v)}{dx} = \frac{du}{dx} - \frac{dv}{dx} \\ \\ \star \qquad\frac{d {x}^{n} }{dx} = n \cdot {x}^{n - 1}$

$\sf \underline{Some \: Formulae} \\ \star \qquad \frac{d \: sin \: x}{dx} = cos \: x \\ \\ \star \qquad \frac{d \: cos \: x}{dx} = - sin \: x \\ \\ \star \qquad \frac{d \: {e}^{x} }{dx} = {e}^{x} \\ \\ \star \qquad \frac{d(nu)}{dx} = n \cdot \frac{du}{dx} \qquad n = \sf constant \\ \\ \star \qquad \frac{d(tan \: x)}{dx} = {sec}^{2} \: x$