Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=t^{2} \sqrt{t-2} $$

Short answer

The derivative of the function \(y=t^{2} \sqrt{t-2}\) is \(y' = 2t * \sqrt{t - 2} + t^{2} * 1/(2\sqrt{t - 2})\).

Step-by-step solution

Identify the two functions

The provided function is a product of two other functions. Hence, we begin by identifying the two functions that make the product. We have: \(f(t)=t^{2}\) and \(g(t)=\sqrt{t-2} = (t-2)^{1/2}\) (by rewriting the square root function in exponential form for easier differentiation).

Apply the Product Rule

Next, apply Product Rule which states that (f.g)' = f'.g + f.g'. This implies \(y' = f'(t)g(t) + f(t)g'(t)\). In this step, we will also need to find \(f'(t)\) and \(g'(t)\), derivatives of \(f(t)\) and \(g(t)\) respectively.

Find the derivative of \(f(t)=t^{2}\)

The derivative of \(f(t)=t^{2}\) is \(f'(t) = 2t\). This differentiation uses the Power Rule which states that the derivative of \(t^n\) is \(n*t^{n-1}\).

Find the derivative of \(g(t) = (t-2)^{1/2}\)

This is a composite function which requires the Chain Rule for differentiation. Chain Rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. Applying the Chain Rule we find, \(g'(t) = 1/2*(t - 2)^{-1/2} * 1\), where 1 is the derivative of the inner function \(t-2\) (using the Constant Rule), and \(-1/2*(t - 2)^{-1/2}\) is the derivative of the outer function \((t-2)^{1/2}\) (using the Power Rule).

Substitute \(f'(t)\), \(f(t)\), \(g'(t)\) and \(g(t)\) into \(y' = f'(t)g(t) + f(t)g'(t)\)

Substitute \(f'(t)\), \(f(t)\), \(g'(t)\) and \(g(t)\) into \(y' = f'(t)g(t) + f(t)g'(t)\). Thus, \(y' = 2t * \sqrt{t - 2} + t^{2} * 1/(2\sqrt{t - 2})\). This is the derivative of the given function.