Question

Let X= 6t^2 – 3t^4 find dx/dt​

Answer 1

Answer:

dx/dt=12t-12t^3=12t(1-t^2)

Step-by-step explanation:

let x=ax^n

so dx/dt=nax^n-1

bybusing above formula

dx/dt=6t^2-3t^4=6*2t-3*4t^3=12t-12t^3=12t(1-t^2)

Answer 2

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$\dfrac{dx}{dt} = 12t(1- {t}^{2} )$

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$\large{➣}$$\dfrac{dx}{dt} = 6 {t}^{2} - 3{t}^{4}$

Differentiating both terms, we get:

$\large{➣}$$\dfrac{dx}{dt} = 6 \times 2 \times {t}^{(2 - 1)} - 3 \times 4 \times {t}^{(4 - 1)}$

Multiplying respective terms, we get:

$\large{➣}$$\dfrac{dx}{dt} = 12t - 12 {t}^{3}$

Taking out 12t as common, we get:

$\large{➣}$$\dfrac{dx}{dt} = 12t(1 - {t}^{2} )$

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$\large{☞}$$\dfrac{dx}{dt} a{x}^{n}$ = $a\times n {x}^{n-1}$

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