Question

✨hєч αll вrαínlч guчѕ✌​.

​ Find the second order derivatives of the function:
${x}^{2} + 3x + 2$
​dσn't ѕpαm✨
fσllσw mє❤​

Answer 1

Answer:

The answer is dy/dx=2x+3

d²y/dx²=2

Answer 2

ANSWER:

$\sf The \ second \ order \ derivative \ of \ {x}^{2} + 3x + 2=2$

GIVEN:

• $\sf{x}^{2} + 3x + 2$

TO FIND:

• $\sf The \ second \ order \ derivative \ of \ {x}^{2} + 3x + 2$

EXPLANATION:

$\sf Let\ y = {x}^{2} + 3x + 2$

Differentiate w.r.t x

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

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

$\boxed {\bold {\large{ \gray{ \dfrac{d}{dx}kx = k }}}}$

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

$\sf \dfrac{dy}{dx} =2x^{2 - 1} + 3 + 0$

$\sf \dfrac{dy}{dx} =2x^{1} + 3$

$\sf \dfrac{dy}{dx} =2x + 3$

Differentiate w.r.t x

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

$\boxed {\bold {\large{ \gray{ \dfrac{d}{dx}kx = k }}}}$

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

$\sf \dfrac{d^{2} y}{dx^{2} } = 2 + 0$

$\sf \dfrac{d^{2} y}{dx^{2} } = 2$

$\sf The \ second \ order \ derivative \ of \ {x}^{2} + 3x + 2= 2$

SOME MORE FORMULAE:

$\bigstar \boxed {\bold {\large{ \orange{ \dfrac{d}{dx}sin \ \theta =cos \ \theta }}}}$

$\bigstar\boxed {\bold {\large{\blue{ \dfrac{d}{dx}cos\ \theta =-sin \ \theta }}}}$

$\bigstar\boxed {\bold {\large{ \green{ \dfrac{d}{dx}tan \ \theta =sec^2 \ \theta }}}}$

$\bigstar\boxed {\bold {\large{ \orange{ \dfrac{d}{dx}sec \ \theta =sec \ \theta \ tan \ \theta }}}}$

$\bigstar\boxed {\bold {\large{ \blue{ \dfrac{d}{dx}cot \ \theta =-cosec^2 \ \theta }}}}$

$\bigstar \boxed {\bold {\large{ \green{ \dfrac{d}{dx}cosec \ \theta =-cosec \ \theta \ cot \ \theta }}}}$