Question

If u = 3t⁴ - 5t³ +2t² - 18t + 4, find du/dt at t= 1.

Answer 1

u = 3t⁴ - 5t³ +2t² - 18t + 4.

=) du/dt = d/dt ( 3t⁴ - 5t³ +2t² - 18t + 4)

= d/dt (3t⁴) - d/dt (5t³) + d/dt (2t²) - d/dt (18t) + d/dt (4)

= 3*4t³ - 5*3t² + 2*2t - 18 + 0

= 12t³ - 15t² + 4t - 18

At t = 1,

= 12(1)³ - 15(1)² + 4(1) - 18

= 12 - 15 + 4 - 18

= - 17.

Answer 2

Given :

$u=3t^4-5t^3+2t^2-18t+4$

$\dfrac{du}{dt}=\dfrac{d}{dt}(3t^4-5t^3+2t^2-18t+4)\\\\\implies \dfrac{du}{dt}=3\times 4t^3-5\times 3t^2+2\times 2t-18\times 1 +0\\\\\implies \dfrac{du}{dt}=12t^3-15t^2+4t-18$

Put t = 1 in the above equation :

$\dfrac{du}{dt}=12-15+4-18\\\\\implies \dfrac{du}{dt}=-3+4-18\\\\\implies \dfrac{du}{dt}=1-18\implies -17$

The value is - 17 .

EXPLANATION :

$\bigstar \dfrac{d}{dx}(3x^4)\implies 3\dfrac{d}{dx}(x^4)\implies 3\times 4x^{4-1}\implies 3 \times 4x^3\\\\\bigstar \dfrac{d}{dx}(5x^3)\implies 5\dfrac{d}{dx}(x^3)\implies 5\times 3x^{3-1}\implies 5 \times 3x^2\\\\\bigstar \dfrac{d}{dx}(2x^2)\implies 2\dfrac{d}{dx}(x^2)\implies 2\times 2x^{2-1}\implies 2 \times 2x$