Question

differenciation of t cube upon 3

Answer 1

Answer:

$t^{2}$

Explanation:

We need to differentiate  $\frac{t^{3} }{3}$, hence we use division rule, which is as follows :

if  $y =\frac{u}{v}$ then,

$\frac{d}{dx}(\frac{u}{v}) = \frac{v.\frac{dx}{dy}(u) - u.\frac{d}{dx}(v) }{v^{2} }$ ; this is also known as quotient rule

Now as per question :

$y = \frac{t^{3} }{2}$ then,

$\frac{d}{dx}(\frac{t^{3} }{3}) = \frac{3.\frac{d}{dx}(t^{3} ) - \frac{t^{3} }{3} .\frac{d}{dx}(3) }{3^{2} }$

we know that :

1. $\frac{d}{dx} (t^{3})$ = $3t^{2}$
2. $\frac{d}{dx} (3)$ = $0$

⇒ $\frac{d}{dx}(\frac{t^{3} }{3}) = \frac{3(3t^{2} ) - \frac{t^{3} }{3} .(0) }{9}$ $= \frac{9t^{2} - 0 }{9}$ $= t^{2}$

Hence,

$\frac{d}{dx}(\frac{t^{3} }{3}) = t^{2}$

Hope it helps :)

(If you find any mistakes, which I am sure there aren't any, then please contact me within 10 minutes of writing this answer, if not possible, please forgive me)

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