Using the differentiation rules and the derivative table, find the derivative of the function y=\sqrt[3]{x}\log_3(x)




y'=\frac{1}{3\sqrt[3]{x^2}\ln(3)}\left(\ln(3)\cdot\log_3(x)+3\right)



Let’s rewrite \sqrt[3]{x} as x^{1/3} so

y = x^{1/3} \log_3(x)


We can use the Product Rule: if y = f(x)\cdot g(x) with f and g differentiable, then the derivative y' is given by:

y' = f'(x)g(x) + f(x)g'(x)

  • Identify the functions


    f(x) = x^{1/3}, \quad g(x) = \log_3(x)


  • Calculate the derivatives of f and g

    These are two elementary functions. Let’s read the derivatives directly from the table:


    f(x) = x^{1/3} \implies f'(x) = \frac{1}{3} x^{-2/3}

    g(x) = \log_3(x) \implies g'(x)= \frac{1}{x\ln(3)}


  • Apply the Product Rule

    Substituting f, g, f' and g' in the formula we find


    y' = \frac{1}{3} x^{-2/3} \cdot \log_3(x) + x^{1/3} \cdot\frac{1}{x\ln(3)}


  • Simplify


    \begin{aligned} y' &= \frac{1}{3} x^{-2/3} \log_3(x) + \frac{1}{\ln(3)} x^{-2/3}\\ &=\frac{1}{3\ln(3)} x^{-2/3}\left(\ln(3)\cdot\log_3(x)+3\right)\\ &= \frac{1}{3\sqrt[3]{x^2}\ln(3)}\left(\ln(3)\cdot\log_3(x)+3\right) \end{aligned} 


    This is the final result, which can also be written as: 

    \frac{1}{3\sqrt[3]{x^2}\ln(3)}\left(\ln(x)+3\right)