 # Calculating Derivative – A function with square roots – Exercise 6273

Exercise

Find the derivative of the following function:

$$f(x)={(\sqrt{x}+\frac{1}{\sqrt{x}})}^{10}$$

$$f'(x)=\frac{5{(x+1)}^9(x-1)}{x^6}$$

Solution

We simplify the function before differentiating:

$$f(x)={(\sqrt{x}+\frac{1}{\sqrt{x}})}^{10}=$$

$$={(\frac{x+1}{\sqrt{x}})}^{10}$$

Using Derivative formulas and the quotient rule in Derivative Rules, we get the derivative:

$$f'(x)=10{(\frac{x+1}{\sqrt{x}})}^9\cdot \frac{\sqrt{x}-(x+1)\cdot\frac{1}{2\sqrt{x}}}{x}=$$

One can simplify the derivative:

$$=10\cdot\frac{{(x+1)}^9}{{(\sqrt{x})}^9}\cdot \frac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=$$

$$=10\cdot\frac{{(x+1)}^9}{{(\sqrt{x})}^9}\cdot\frac{2x-x-1}{2x\sqrt{x}}=$$

$$=\frac{5{(x+1)}^9(x-1)}{x^{\frac{9}{2}}\cdot x\cdot x^{\frac{1}{2}}}=$$

$$=\frac{5{(x+1)}^9(x-1)}{x^6}$$

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