Question

differentiation solve : 
y=(x−1)(x2+x+1) 
please​ solve on book​

Answer 1

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

$\red{\rm :\longmapsto\:y = (x - 1)( {x}^{2} + x + 1)}$

$\rm :\longmapsto\:y = (x - 1)( {x}^{2} + x \times 1 + {1}^{2})$

We know,

$\purple{\boxed{ \sf{ \: (x - y)( {x}^{2} + xy + {y}^{2}) = {x}^{3} + {y}^{3}}}}$

$\rm :\longmapsto\:y = {x}^{3} - {1}^{3}$

$\rm :\longmapsto\:y = {x}^{3} - {1}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx}({x}^{3} - {1} )$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx}{x}^{3} - \dfrac{d}{dx}{1}$

We know,

$\: \: \: \: \: \: \: \: \boxed{ \sf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}} \: \: and \: \: \boxed{ \sf{ \: \dfrac{d}{dx}k = 0}}$

Using these, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{3 - 1} - 0$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{2}$

Additional Information :-

$\red{\rm :\longmapsto\:\dfrac{d}{dx}x = 1}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {e}^{x} = {e}^{x} }$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}logx = \dfrac{1}{x}}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}sinx = cosx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosx = - \: sinx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosecx = - \: cosecx \: cotx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}secx = \: secx \: tanx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}tanx = {sec}^{2}x}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = { - \: cosec}^{2}x}$

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)