Question

if x=t^3+1/t2 then dx/dt is? ​

Answer 1

Solution:-

We have

$\rm \: if \: x = {t}^{3} + \dfrac{1}{ {t}^{2} } \: then \: \: \dfrac{dx}{dt}$

Now we can write as

$\rm \: x = {t}^{3} + {t}^{ - 2}$

Using algebra of differentiation

$\boxed{ \rm \: \dfrac{d}{dx} (f(x) \pm \: g(x)) = \dfrac{d}{dx} f(x) \pm \dfrac{d}{dx} g(x)}$

Now we can write

$\rm \: x = {t}^{3} + {t}^{ - 2}$

$\rm \dfrac{d}{dx} {t}^{3} + \dfrac{d}{dx} {t}^{ - 2}$

Using standard derivation

$\rm\dfrac{d}{dx} ( {x}^{n} ) = nx {}^{n - 1}$

we get

$\rm \: 3t {}^{3 - 1} - 2t {}^{ - 2 - 1}$

$\rm \: 3 {t}^{2} - 2t {}^{ - 3}$

$\to \rm \: 3{t}^{2} - \dfrac{2}{ {t}^{3} }$

Option 2 is correct

Answer 2

Answer:

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Explanation:

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