Question

It is question from differentiation.
Pls answer it

Answer 1

Answer:-

$v = 20t^4 + 9t^2 + 9 - \dfrac{5}{t^2}$

Given :-

$x = 4t^5 +3t^3 +9t + \dfrac{5}{t}$

where,

where, x is displacement.

To find :-

$\dfrac{dx}{dt}$

Solution:-

→$x = 4t^5 +3t^3 +9t + \dfrac{5}{t}$

→$x = 4t^5 + 3t^3 + 9t+ 5t^{-1}$

• Differentiate with respect to time.

→$\dfrac{dx}{dt}= \dfrac{d(4t^5+3t^3+9t+ 5t^{-1})}{dt}$

→$\dfrac{dx}{dt}= 4\dfrac{d(t^5)}{dt}+ 3\dfrac{d(t^3)}{dt}+9\dfrac{dt}{dt} + 5 \dfrac{d(t^{-1})}{dt}$

• The differentiation of displacement with respect to time gives velocity.

→$v = 4 \times 5 t^4 + 3 \times 3 t^2 + 9 \times 1 + 5 \times -1t^{-2}$

→$v = 20t^4 +9t^2 + 9 -5t^{-2}$

→$v = 20t^4 + 9t^2 + 9 - \dfrac{5}{t^2}$

Answer 2

something personal happened cannot tell you

sorry dear .......

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