Find the derivative for
$$(\sqrt{3x} - \sqrt[3]{x})^3 (x+3)^{1/3}.$$
I'd tried to use the formula $f'(x) = u'(x)v(x) + v'(x)u(x)$, but it become very complicated. It will produce many fraction and I cannot solve for that.
Differentiation of products
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$\begingroup$
calculus
derivatives
products
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1 Answers
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Hint
Logarithmic differentiation would make life easier $$y=(\sqrt{3x} - \sqrt[3]{x})^3 (x+3)^{1/3}$$ $$\log(y)=3\log(\sqrt{3x} - \sqrt[3]{x})+\frac 13\log(x+3)$$ Differentiate $$\frac{y'}y=3 \frac{\frac{\sqrt{3}}{2 \sqrt{x}}-\frac{1}{3 x^{2/3}}}{\sqrt{3} \sqrt{x}-\sqrt[3]{x}}+\frac 13 \frac 1 {x+3}$$ Simplify and, later, $y'=y\times\frac{y'}y$.