S(t) = 0.05t^3 + 0.3t^2 + 2t + 8
Find S'(t)
just straightforward polynomial.
S' = 0.15t^2 + 0.6t + 2
Something tells me the decimals scared you off. Don't forget your 4th-grade math just because you're learning calculus!
To find the derivative of the function S(t), denoted as S'(t), we will use the power rule and apply it term by term.
The power rule states that if we have a term of the form k * x^n, where k is a constant, and n is any real number, then the derivative of that term is given by:
d/dx (k * x^n) = k * n * x^(n-1)
Let's differentiate each term of the function S(t) individually:
Term 1: 0.05t^3
Applying the power rule, the derivative is:
d/dt (0.05t^3) = 0.05 * 3 * t^(3-1) = 0.15t^2
Term 2: 0.3t^2
Applying the power rule, the derivative is:
d/dt (0.3t^2) = 0.3 * 2 * t^(2-1) = 0.6t
Term 3: 2t
The derivative of 2t is simply 2, as the derivative of a constant multiplied by t is the constant itself.
Term 4: 8
The derivative of a constant term is zero, as the rate of change of a constant is always zero.
Now let's combine these derivatives to find S'(t):
S'(t) = 0.15t^2 + 0.6t + 2
Therefore, the derivative of S(t) is S'(t) = 0.15t^2 + 0.6t + 2.