Question

Differentiation of 3x²+4x+3 with respect to x is​

Answer 1

ANSWER:

• Differentiation of 3x² + 4x + 3 = 6x + 4

GIVEN:

• 3x² + 4x + 3

TO DIFFERENTIATE:

• 3x² + 4x + 3 with respect to x.

EXPLANATION:

$\sf \dashrightarrow \dfrac{d}{dx} (3 {x}^{2} + 4x + 3)$

$\boxed{ \bold{ \large{ \pink{ \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}}}}}$

$\boxed{ \bold{ \large{ \blue{ \dfrac{d}{dx} k = 0}}}}$

$\sf \dashrightarrow 2(3 {x}^{2 - 1}) + 1(4 {x}^{1 - 1}) + 0$

$\sf \dashrightarrow 6 {x}^{1} + 4 {x}^{0}$

$\boxed{ \bold{ \large{ \red{ {a}^{0} = 1}}}}$

$\sf \dashrightarrow 6x+ 4$

$\sf \dashrightarrow \dfrac{d}{dx} (3 {x}^{2} + 4x + 3) = 6x + 4$

Hence the differentiation of 3x² + 4x + 3 = 6x + 4.

EXTRA CALCULATION:

TO INTEGRATE:

• 3x² + 4x + 3

EXPLANATION:

$\displaystyle\sf \longmapsto \int (3 {x}^{2} + 4x + 3)$

$\boxed{ \bold{ \large{ \orange{ \int {x}^{n} = \dfrac{{x}^{n +1}}{x+1}+C}}}}$

$\sf \longmapsto 3\bigg(\dfrac{x^{2+1}}{2+1}\bigg)+4\bigg(\dfrac{x^{1+1}}{1+1}\bigg)+3\bigg(\dfrac{x^{0+1}}{0+1}\bigg)+C$

$\sf \longmapsto 3\bigg(\dfrac{x^{3}}{3}\bigg)+4\bigg(\dfrac{x^{2}}{2}\bigg)+3\bigg(\dfrac{x^{1}}{1}\bigg)+C$

$\sf \longmapsto x^3+2x^2+3x+C$

$\displaystyle\sf \longmapsto \int (3 {x}^{2} + 4x + 3) = x^3+2x^2+3x+C$

Hence the integration of 3x² + 4x + 3 = x³ + 2x² +3 x + C.