Find the derivative of the function x=1/2t^4-5t-3

Question

Find the derivative of the function x=1/2t^4-5t-3 USING THE LONG METHOD

Answer 1

I'm going to make the assumption that "USING THE LONG METHOD" means using the limit definition of the derivative and that you are looking for the derivative of "x" with respect the "t".

So dx/dt (also known as x') = limit (as h approaches zero) of (f(t + h) - f(t))/h

where f(t + h) = 1/2(t + h)⁴ - 5(t + h) - 3

f(t + h) = 1/2(t + h)²(t + h)² - 5(t + h) - 3

f(t + h) = 1/2(t² + 2th + h²)(t² + 2th + h²) - 5t - 5h - 3

f(t + h) = 1/2(t⁴ + 4t³h + 6t²h² + 4th3 + h⁴) - 5t - 5h - 3

f(t + h) = 1/2t⁴ + 2t³h + 3t²h² + 2th3 + 1/2h⁴ - 5t - 5h - 3

So f(t + h) - f(t) = 1/2t⁴ + 2t³h + 3t²h² + 2th³ + 1/2h⁴ - 5t - 5h - 3 - (1/2t⁴ - 5t - 3)

f(t + h) - f(t) = 1/2t⁴ + 2t³h + 3t²h² + 2th³ + 1/2h⁴ - 5t - 5h - 3 - 1/2t⁴ + 5t + 3

f(t + h) - f(t) = 2t³h + 3t²h² + 2th³ + 1/2h⁴ - 5h

f(t + h) - f(t) = h(2t³ + 3t²h + 2th² + 1/2h³ - 5)

And [f(t + h) - f(t)]/h = h(2t³ + 3t²h + 2th² + 1/2h³ - 5)/h = 2t³ + 3t²h + 2th² + 1/2h³ - 5

The limit as h approaches zero is 2t³ - 5

So x' = 2t³ - 5