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### Section 3-4 : Product and Quotient Rule

2. Use the Product Rule to find the derivative of $$y = \left( {1 + \sqrt {{x^3}} } \right)\,\left( {{x^{ - 3}} - 2\sqrt[3]{x}} \right)$$ .

Show Solution

There isn’t much to do here other than take the derivative using the product rule. We’ll also need to convert the roots to fractional exponents.

$y = \left( {1 + {x^{\frac{3}{2}}}} \right)\,\left( {{x^{ - 3}} - 2{x^{\frac{1}{3}}}} \right)$

The derivative is then,

$\frac{{dy}}{{dx}} = \left( {\frac{3}{2}{x^{\frac{1}{2}}}} \right)\,\left( {{x^{ - 3}} - 2{x^{\frac{1}{3}}}} \right) + \left( {1 + {x^{\frac{3}{2}}}} \right)\,\left( { - 3{x^{ - 4}} - \frac{2}{3}{x^{ - \,\,\frac{2}{3}}}} \right) = - 3{x^{ - 4}} - \frac{3}{2}{x^{ - \,\,\frac{5}{2}}} - \frac{2}{3}{x^{ - \,\,\frac{2}{3}}} - \frac{{11}}{3}{x^{\frac{5}{6}}}$

Note that we multiplied everything out to get a “simpler” answer.