Question

if x=10t^2-10/t plus 20
find dx/dt​

Answer 1

$\mathbb{\underline {ANSWER}}$

$\frac{dx}{dt} = 20t +\frac{10}{t^{2} }$

$\mathbb {\underline {EXPLANATION}}$

$x = 10t^{2}-\frac{10}{t} +20$

$\frac{dx}{dt} =\frac{d}{dt} [10t^{2}-\frac{10}{t} +20 ]$

$\frac{dx}{dt} = \frac{d}{dt} (10t^{2} ) + \frac{d}{dt} (\frac{-10}{t} )+ \frac{d}{dt} (20)$

We know that,

$\frac{d}{dt} (t^{n} )=nt^{n-1}$

$\frac{d}{dt} (kx)=k \frac{d}{dt} (x)$   where k is a constant

$\frac{d}{dt} (k) = 0$             where k is a constant

Hence,

$\frac{d}{dt} (10t^{2} ) = 10* \frac{d}{dt} (t^{2} )=10*2t = 20t$

$\frac{d}{dt} (\frac{-10}{t} )= \frac{d}{dt} (-10t^{-1} )=-10* \frac{d}{dt} (t^{-1} )=-10*-1*t^{-1-1}= \frac{10}{t^{2} }$

$\frac{d}{dt} (20)=0$

Therefore,

$\frac{dx}{dt} = \frac{d}{dt} (10t^{2} )+ \frac{d}{dt} (\frac{-10}{t} )+ \frac{d}{dt} (20)$

$\Rightarrow$ $\frac{dx}{dt} = 20t +\frac{10}{t^{2} }$