Question

in dx/dt where x=a+bt², how did 2bt come on solving dx/dt?

Answer 1

Steps:
1)
We have,

$x = a + b {t}^{2}$
where a, b are constants.

2)Then,

$\frac{d(x)}{dt} = \frac{d(a)}{dt} + \frac{d(b {t}^{2} )}{dt}$

We know that, differentiation of constant function is 0 .
Here,
( a ) is constant.
So,
$\frac{d(a)}{dt} = 0$

3) After that, we know

$\frac{d( {x}^{n} )}{dx} = n {x}^{(n - 1)}$
where n is constant.
Therefore,
$\frac{d(b {t}^{2}) }{dt} = b \: \times \frac{d( {t}^{2} )}{dt} = b \times (2 {t}^{(2 - 1)} ) \\ = b \times (2t)$

4) Therefore,
$\frac{d(x)}{dt} = 0 + 2bt \: = 2bt$