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Chain Rule of Differentiation

Find the deri...

Question

Find the derivative of $y = \frac{1}{x} + \frac{1}{\sqrt{x}} + \frac{1}{\sqrt[3]{x}}$

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Solution

$y = \frac{1}{x} + \frac{1}{\sqrt{x}} + \frac{1}{\sqrt[3]{x}}$

So, $\frac{d y}{d x} = ?$

$y = x^{- 1} + x^{\frac{- 1}{2}} + x^{\frac{- 1}{3}}$

$\frac{d y}{d x} = - 1 x^{- 2} + \frac{1}{2} x^{\frac{- 1}{2} - 1} - \frac{- 1}{3} x^{\frac{- 1}{3} - 1} [\frac{d}{d x} (x^{n}) = n x^{n - 1}]$

$= \frac{- 1}{x^{2}} - \frac{- 1}{2} x^{\frac{- 3}{2}} - \frac{1}{3} x^{\frac{- 1}{3}}$

$= \frac{- 1}{x^{2}} - \frac{1}{2 x^{\frac{3}{2}}} - \frac{1}{3 x^{\frac{4}{3}}}$

$= \frac{- 1}{x^{2}} - \frac{1}{2 x \sqrt{x}} - \frac{1}{3 x^{1} \cdot x^{\frac{1}{3}}}$

$= \frac{- 1}{x^{2}} - \frac{2 x \sqrt{x}}{-} \frac{1}{3 x x^{\frac{1}{3}}}$

$= - \frac{(6 + 3 \sqrt{x} + 2 \sqrt[3]{x^{3}})}{6 x^{2}}$ .

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