[tex]\large\underline{\sf{Solution-}}[/tex] [tex]\red{\rm :\longmapsto\:y = (x – 1)( {x}^{2} + x + 1)}[/tex] [tex]\rm :\longmapsto\:y = (x – 1)( {x}^{2} + x \times 1 + {1}^{2})[/tex] We know, [tex] \purple{\boxed{ \sf{ \: (x – y)( {x}^{2} + xy + {y}^{2}) = {x}^{3} + {y}^{3}}}}[/tex] [tex]\rm :\longmapsto\:y = {x}^{3} – {1}^{3} [/tex] [tex]\rm :\longmapsto\:y = {x}^{3} – {1} [/tex] On differentiating both sides w. r. t. x, we get [tex]\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx}({x}^{3} – {1} )[/tex] [tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx}{x}^{3} – \dfrac{d}{dx}{1}[/tex] We know, [tex] \: \: \: \: \: \: \: \: \boxed{ \sf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n – 1}}} \: \: and \: \: \boxed{ \sf{ \: \dfrac{d}{dx}k = 0}}[/tex] Using these, we get [tex]\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{3 – 1} – 0[/tex] [tex]\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{2} [/tex] Additional Information :- [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}x = 1}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx} {e}^{x} = {e}^{x} }[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}logx = \dfrac{1}{x}}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}sinx = cosx}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}cosx = – \: sinx}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}cosecx = – \: cosecx \: cotx}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}secx = \: secx \: tanx}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}tanx = {sec}^{2}x}[/tex] [tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = { – \: cosec}^{2}x}[/tex] 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) Reply
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\red{\rm :\longmapsto\:y = (x – 1)( {x}^{2} + x + 1)}[/tex]
[tex]\rm :\longmapsto\:y = (x – 1)( {x}^{2} + x \times 1 + {1}^{2})[/tex]
We know,
[tex] \purple{\boxed{ \sf{ \: (x – y)( {x}^{2} + xy + {y}^{2}) = {x}^{3} + {y}^{3}}}}[/tex]
[tex]\rm :\longmapsto\:y = {x}^{3} – {1}^{3} [/tex]
[tex]\rm :\longmapsto\:y = {x}^{3} – {1} [/tex]
On differentiating both sides w. r. t. x, we get
[tex]\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx}({x}^{3} – {1} )[/tex]
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx}{x}^{3} – \dfrac{d}{dx}{1}[/tex]
We know,
[tex] \: \: \: \: \: \: \: \: \boxed{ \sf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n – 1}}} \: \: and \: \: \boxed{ \sf{ \: \dfrac{d}{dx}k = 0}}[/tex]
Using these, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{3 – 1} – 0[/tex]
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{2} [/tex]
Additional Information :-
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}x = 1}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx} {e}^{x} = {e}^{x} }[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}logx = \dfrac{1}{x}}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}sinx = cosx}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}cosx = – \: sinx}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}cosecx = – \: cosecx \: cotx}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}secx = \: secx \: tanx}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}tanx = {sec}^{2}x}[/tex]
[tex]\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = { – \: cosec}^{2}x}[/tex]
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)