Question

differentiate the function.
\\( u = sqrt3{t} + 2sqrt{t^3} \\)
\\( u = \\) \\(\times\\)

Explanation:

Step1: Rewrite the terms with exponents

Rewrite \(\sqrt[3]{t}\) as \(t^{\frac{1}{3}}\) and \(\sqrt{t^3}\) as \(t^{\frac{3}{2}}\). So the function becomes \(u = t^{\frac{1}{3}}+ 2t^{\frac{3}{2}}\).

Step2: Apply the power rule for differentiation

The power rule states that if \(y = t^n\), then \(y'=nt^{n - 1}\).
For the first term \(t^{\frac{1}{3}}\), using the power rule, the derivative is \(\frac{1}{3}t^{\frac{1}{3}-1}=\frac{1}{3}t^{-\frac{2}{3}}\).
For the second term \(2t^{\frac{3}{2}}\), using the power rule, the derivative is \(2\times\frac{3}{2}t^{\frac{3}{2}-1}=3t^{\frac{1}{2}}\).

Step3: Combine the derivatives

Combine the derivatives of the two terms. So \(u'=\frac{1}{3}t^{-\frac{2}{3}}+3t^{\frac{1}{2}}\). We can also rewrite \(t^{-\frac{2}{3}}\) as \(\frac{1}{t^{\frac{2}{3}}}=\frac{1}{\sqrt[3]{t^{2}}}\) and \(t^{\frac{1}{2}}\) as \(\sqrt{t}\). So \(u'=\frac{1}{3\sqrt[3]{t^{2}}}+3\sqrt{t}\) or in terms of exponents \(u'=\frac{1}{3}t^{-\frac{2}{3}} + 3t^{\frac{1}{2}}\).

Answer:

\(\frac{1}{3}t^{-\frac{2}{3}}+3t^{\frac{1}{2}}\) (or equivalent forms like \(\frac{1}{3\sqrt[3]{t^2}} + 3\sqrt{t}\))