Question

If y = x cosecx , then find dy/dx ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: y = x \: cosecx$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}(x \: cosecx)$

We know,

$\boxed{ \rm{ \:\dfrac{d}{dx}(u.v) = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

So, using this result, we get

$\rm \: \dfrac{dy}{dx} = x\dfrac{d}{dx}cosecx + cosecx\dfrac{d}{dx}x$

$\rm \: \dfrac{dy}{dx} = x( - cosecx \: cotx) + cosecx \times 1$

$\rm \: \dfrac{dy}{dx} = - x \: cosecx \: cotx \: + \: cosecx$

$\bf\implies \:\dfrac{dy}{dx} = cosecx( - x \: cotx \: + \: 1)$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$