Question

find the derivative of the following function.
f(t) = t^{\frac{8}{5}}e^{t}
f(t) = square

Explanation:

Step1: Apply product - rule

The product - rule states that if \(y = u\cdot v\), then \(y'=u'v + uv'\). Let \(u = t^{\frac{8}{5}}\) and \(v = e^{t}\).

Step2: Find the derivative of \(u\)

Using the power - rule \((x^{n})'=nx^{n - 1}\), for \(u=t^{\frac{8}{5}}\), we have \(u'=\frac{8}{5}t^{\frac{8}{5}-1}=\frac{8}{5}t^{\frac{3}{5}}\).

Step3: Find the derivative of \(v\)

The derivative of \(v = e^{t}\) is \(v'=e^{t}\).

Step4: Calculate \(f'(t)\)

By the product - rule \(f'(t)=u'v+uv'\), substituting \(u\), \(u'\), \(v\), and \(v'\) we get \(f'(t)=\frac{8}{5}t^{\frac{3}{5}}e^{t}+t^{\frac{8}{5}}e^{t}=t^{\frac{3}{5}}e^{t}(\frac{8}{5}+t)\).

Answer:

\(t^{\frac{3}{5}}e^{t}(\frac{8}{5}+t)\)