Question

find the derivative with respect to t, of function X=A0+A1t+A2t² where A0, A1, A2 are constants plz answer​

Answer 1

Given:

Function is provided as :

$\boxed{x = A_{0} + A_{1}t + A_{2} {t}^{2} }$

To find:

Derivative of this function with respect to "t".

Calculation:

$\therefore \: x = A_{0} + A_{1}t + A_{2} {t}^{2}$

$= > \: \dfrac{dx}{dt} = \dfrac{d( A_{0} + A_{1}t + A_{2} {t}^{2}) }{t}$

Applying Addition rule of Differentiation:

$= > \: \dfrac{dx}{dt} = \dfrac{d(A_{0})}{dt} + \dfrac{d(A_{1}t)}{dt} + \dfrac{d(A_{2} {t}^{2}) }{dt}$

$= > \: \dfrac{dx}{dt} = \dfrac{d(A_{0})}{dt} + A_{1}\dfrac{d(t)}{dt} + A_{2}\dfrac{d( {t}^{2}) }{dt}$

$= > \: \dfrac{dx}{dt} = 0 + A_{1}(1)+ A_{2}(2t)$

$= > \: \dfrac{dx}{dt} = 0 + A_{1}+ 2A_{2}t$

$= > \: \dfrac{dx}{dt} = A_{1}+ 2A_{2}t$

So, final answer is:

$\boxed{ \bold{ \: \dfrac{dx}{dt} = A_{1}+ 2A_{2}t}}$