Question

( tanx + secx) ( cotx+ cosecx)differentiate it​

Answer 1

Given :

$\\ \bullet\bf{Differentiation \: of \: the \: following \: expression }$

$\\ \red\bigstar\bf{(tanx+secx)(cotx+cosecx)}$

To Find :

$\\ \bullet\bf{Differentiation \: of \: the \: following \: equation =?}$

Solution :

• Differentiating both sides with respect to x x :

$\\ \boxed{\blue{\bf{By \: using \: Product \: Rule}}}$

$\\ \implies \sf \: y = (tanx + secx).(cotx + cosecx) \\ \\ \\ \implies \sf \: \frac{dy}{dx} = (tanx + secx) \frac{d}{dx} \bigg (cotx + cosecx \bigg) + (cotx + cosecx) \frac{d}{dx} \bigg(tanx + secx \bigg) \\ \\ \\ \implies \sf \: \frac{dy}{dx} = (tanx + secx) \bigg( - {cosec}^{2} x - cotx.cosecx \bigg) + (cotx + cosecx) \bigg( {secx}^{2} + secx.tanx \bigg) \\ \\ \\ \implies \sf \: \frac{dy}{dx} = (tanx + secx) \bigg( - {cosec}^{2} x - cotx.tanx \bigg) + (cotx + cosecx) \bigg( {sec}^{2} x + secx.tanx \bigg) \\ \\ \\ \implies \sf \: \frac{dy}{dx} = (cotx + cosecx)(secx + tanx)secx - cosecx(tanx + secx)(cotx + cosecx) \\ \\ \\ \implies \sf \: \frac{dy}{dx} = (cotx + cosecx)(secx + tanx)(secx - cosecx) \\ \\ \\ \implies \boxed{ \sf{ \frac{dy}{dx} = (cotx + cosecx)(secx + tanx)(secx - cosecx}} \pink \bigstar$

$\green{\underline{\sf{Differentiation \: of \: (tanx+secx)(cotx+cosecx) \: is}}}$

$\red \bigstar \boxed{ \sf{ \frac{dy}{dx} = (cotx + cosecx)(secx + tanx)(secx - cosecx}}$

Formulas :

$\bf \: (1) \: \frac{d}{dx} sinx = cosx$

$\\ \bf \: (2) \: \frac{d}{dx} cosx = - sinx$

$\\ \bf \: (3) \frac{d}{dx} tanx = {sec}^{2} x$

$\\ \bf \: (4) \frac{d}{dx} cotx = - {cosec}^{2} x$

$\\ \bf \: (5) \: \frac{d}{dx} secx = secx.tanx$

$\\ \bf \: (6) \: \frac{d}{dx} cosecx = - cosecx.cotx$

$\\ \bf \: (7) \: \frac{d}{dx} logx = \frac{1}{x}$

$\\ \bf \: (8) \frac{d}{dx} {x}^{n} = n {x}^{n - 1}$

$\\ \bf \: (9) \: \frac{d}{dx} cu = c \frac{du}{dx}$

$\\ \bf \: (10) \: \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx} .... \red{ \bf{ (\: product \: rule)}}$

$\\ \bf \: (11) \frac{d}{dx} \frac{u}{v} = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {v}^{2} } \: \: .... \red{ \bf{(quotient \: rule)}}$