Derivative Problem

Question

Let f(t)= √2/t³

Find f'(t)

Find f'(3)

Answer 1

f(x) = sqr2/t^3 = sqr2(t^-3)

f'(x) =-3sqr2(t^-4) = -(3sqr2)/t^4

f'(3) =-3sqr2/81= -(sqr2)/27

f'(3) = about -0.052

Answer 2

Rewrite f(t) as f(t) = √2 t^-3

Now apply the power rule:

f'(t) = -3√2 t^-4 = -3√2 / t⁴

So the slope of the tangent line at the point where x = 3 is -3√2/81 = -√2/27

Here is a graph showing the tangent line at the point (3, √2/27). This graph also has a slider for points on the graph of f and shows the tangent line at points where the slider is located.

desmos.com/calculator/7bn0u34ydv