How do you differentiate $\sqrt[3]{x} + \sqrt[4]{x}$ $root3(x)+root4(x)$ ?

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How do you differentiate $\sqrt[3]{x} + \sqrt[4]{x}$ $root3(x)+root4(x)$ ?

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Answer 1 · 2016-08-08T17:56:19

$= \frac{1}{3 x^{\frac{2}{3}}} + \frac{1}{4 x^{\frac{3}{4}}}$ $=1/(3x^(2/3))+1/(4x^(3/4))$ .

Explanation:

We know that, $\frac{d}{d x} (x^{n}) = n \cdot x^{n - 1}$ $d/dx(x^n)=n*x^(n-1)$ , and,

$\frac{d}{d x} (u + v) = \frac{d u}{d x} + \frac{d v}{d x}$ $d/dx(u+v)=(du)/dx+(dv)/dx$

Therefore, $\frac{d}{d x} (\sqrt[3]{x} + \sqrt[4]{x}) = \frac{d}{d x} \sqrt[3]{x} + \frac{d}{d x} \sqrt[4]{x}$ $d/dx(root3x+root4x)=d/dxroot3x+d/dxroot4x$

$= \frac{d}{d x} x^{\frac{1}{3}} + \frac{d}{d x} x^{\frac{1}{4}}$ $=d/dxx^(1/3)+d/dxx^(1/4)$

$= \frac{1}{3} \cdot x^{\frac{1}{3} - 1} + \frac{1}{4} \cdot x^{\frac{1}{4} - 1}$ $=1/3*x^(1/3-1)+1/4*x^(1/4-1)$

$= \frac{1}{3 x^{\frac{2}{3}}} + \frac{1}{4 x^{\frac{3}{4}}}$ $=1/(3x^(2/3))+1/(4x^(3/4))$ .

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How do you differentiate $\sqrt[3]{x} + \sqrt[4]{x}$ $root3(x)+root4(x)$ ?

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