Question

Use the limit definition to find the derivative of the function. $$ f(t)=t^{3}+t^{2} $$

Short answer

The derivative of the function \(f(t) = t^3 + t^2\) using the limit definition is \(f'(t) = 3t^2 + 2t\).

Step-by-step solution

Write the limit definition of derivative

The limit definition of the derivative \(f'(t)\) is given by \[ f'(t) = \lim_{h \to 0} \frac{f(t + h) - f(t)}{h} \]

Substitute the given function

Now, substitute the given function \( f(t) = t^3 + t^2 \) into the limit definition in place of \( f(t) \) and \( f(t + h) \): \[ f'(t) = \lim_{h \to 0} \frac{[(t+h)^3 + (t+h)^2] - [t^3 + t^2]}{h} \]

Expand and simplify the equation

Next, expand the terms in the numerator, simplify them and group like terms: \[ f'(t) = \lim_{h \to 0} \frac{3t^2h + 3th^2 + h^3 + 2th + h^2 - t^3 - t^2 }{h} \] Simplify by canceling out terms that appear on both sides of the subtraction sign in the numerator: \[ f'(t) =\lim_{h \to 0} \frac{3t^2h + 3th^2 + h^3 + 2th + h^2 }{h} \]

Factor out common term

Factoring out the common term \(h\) from the numerator we have: \[ f'(t) = \lim_{h \to 0} h[3t^2 + 3th + h + 2t + h ] \] Divide the term \(h\) in both the numerator and denominator: \[ f'(t) = \lim_{h \to 0} [3t^2 + 3th + h + 2t + h ] \]

Compute the limit

At this point, we can now compute the limit by substituting \(h = 0\) into the expression: \[ f'(t) = 3t^2 + 2t \]