Question:
Find the derivative of $ \sqrt{2}x+2\sqrt{x}-\frac{1}{x} $ ?
Find the derivative of $ \sqrt{2}x+2\sqrt{x}-\frac{1}{x} $ ?
Updated On: Jun 23, 2024
- $ \sqrt{2}+1/\sqrt{x}(1- \frac {1}{2} x) $
- $ \sqrt{2}-1/\sqrt{x}(1+\frac {1}{2}x) $
- $ \sqrt{2}-1/\sqrt{x}(1-\frac {1}{2} x) $
- $ \sqrt{2}+1/\sqrt{x}(1+\frac {1}{2} x) $
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The Correct Option is D
Solution and Explanation
The derivative of $ \sqrt{2}x+2\sqrt{x}-\frac{1}{\sqrt{x}} $
$ =\sqrt{2}+2\cdot \frac{1}{2}{{x}^{\frac{1}{2}-1}}+\frac{1}{2}{{x}^{-\frac{1}{2}-1}} $
$ =\sqrt{2}+{{x}^{\frac{-1}{2}}}+\frac{1}{2}{{x}^{\frac{-3}{2}}} $
$ =\sqrt{2}+\frac{1}{\sqrt{x}}+\frac{1}{2x\sqrt{x}} $
$ =\sqrt{2}+\frac{1}{\sqrt{x}}\left( 1+\frac{1}{2x} \right) $
$ =\sqrt{2}+2\cdot \frac{1}{2}{{x}^{\frac{1}{2}-1}}+\frac{1}{2}{{x}^{-\frac{1}{2}-1}} $
$ =\sqrt{2}+{{x}^{\frac{-1}{2}}}+\frac{1}{2}{{x}^{\frac{-3}{2}}} $
$ =\sqrt{2}+\frac{1}{\sqrt{x}}+\frac{1}{2x\sqrt{x}} $
$ =\sqrt{2}+\frac{1}{\sqrt{x}}\left( 1+\frac{1}{2x} \right) $
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Concepts Used:
Limits And Derivatives
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.
Limits Formula:
Derivatives of a Function:
A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.
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