Question

please solve it .its differentiation sum​

Answer 1

d/dt(6t²-5t^8+10)

==> 6(2)t^2-1-5(8)t^8-1+10(0)

===> 12t-40t^7+0

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$<marquee direction= "left"> 12t-40t^7 is the answer.$

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Answer 2

Hey there !

Just follow the formula :

$\dfrac{d}{dx}\, x^{n} = n\, x^{n-1}$

According to linearity of differentiation:

$\dfrac{d}{dx}\, (ax^{p} + bx^{q}+c)= \dfrac{d}{dx}\, (ax^{p}) + \dfrac{d}{dx}\, (bx^{q})+ \dfrac{d}{dx}\, (c)$

=> $\dfrac{d}{dt}\, (6t^{2} - 5t^{8} + 10)= \dfrac{d}{dt}\, (6t^{2}) - \dfrac{d}{dt}\, (5t^{8})+ \dfrac{d}{dx}\, (10)$

$\dfrac{d}{dt}\, t^{2} = 2\, t^{2-1}= 2t$

$\dfrac{d}{dt}\, t^{8} = 8\, t^{8-1} = 8\, t^{7}$

$\dfrac{d}{dt}\, 10= 0$

=> $\dfrac{d}{dt}\, (6t^{2} - 5t^{8} + 10)= 6(2t) - 5(8\, t^{7}) + 0$

=> $\dfrac{d}{dt}\, (6t^{2} - 5t^{8} + 10) = 12t - 40t^{7}$

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