# How do you differentiate #root3(x)+root4(x)#?

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To differentiate ( \sqrt{3}(x) + \sqrt{4}(x) ), we need to use the power rule of differentiation.

The power rule states that the derivative of ( x^n ) with respect to ( x ) is ( nx^{n-1} ), where ( n ) is a constant.

For ( \sqrt{3}(x) ), we can rewrite it as ( 3^{1/2}x ), and for ( \sqrt{4}(x) ), we can rewrite it as ( 4^{1/2}x ).

Therefore, the derivative of ( \sqrt{3}(x) + \sqrt{4}(x) ) is:

[ \frac{d}{dx} \left( \sqrt{3}(x) + \sqrt{4}(x) \right) = \frac{d}{dx} \left( 3^{1/2}x + 4^{1/2}x \right) ]

[ = \frac{d}{dx} \left( 3^{1/2}x \right) + \frac{d}{dx} \left( 4^{1/2}x \right) ]

[ = \frac{1}{2} \cdot 3^{1/2}x^{(1/2) - 1} + \frac{1}{2} \cdot 4^{1/2}x^{(1/2) - 1} ]

[ = \frac{1}{2} \cdot 3^{1/2}x^{-1/2} + \frac{1}{2} \cdot 4^{1/2}x^{-1/2} ]

[ = \frac{1}{2} \cdot \frac{3^{1/2}}{x^{1/2}} + \frac{1}{2} \cdot \frac{4^{1/2}}{x^{1/2}} ]

[ = \frac{1}{2} \cdot \frac{\sqrt{3}}{\sqrt{x}} + \frac{1}{2} \cdot \frac{\sqrt{4}}{\sqrt{x}} ]

[ = \frac{1}{2} \cdot \frac{\sqrt{3}}{\sqrt{x}} + \frac{1}{2} \cdot \frac{2}{\sqrt{x}} ]

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

So, the derivative of ( \sqrt{3}(x) + \sqrt{4}(x) ) is ( \frac{1}{2} \cdot \left( \frac{\sqrt{3} + 2}{\sqrt{x}} \right) ).

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