 # Calculating Derivative – A function with ln and square roots – Exercise 6271

Exercise

Find the derivative of the following function:

$$f(x)=\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\ln(x+\sqrt{x^2-9})$$

$$f'(x)=\sqrt{x^2-9}$$

Solution

$$f(x)=\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\ln(x+\sqrt{x^2-9})$$

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

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

One can simplify the derivative:

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

$$=\frac{\sqrt{x^2-9}}{2}+\frac{x^2}{2\sqrt{x^2-9}}-\frac{9}{2\sqrt{x^2-9}}=$$

$$=\frac{x^2-9+x^2-9}{2\sqrt{x^2-9}}=$$

$$=\frac{2(x^2-9)}{2\sqrt{x^2-9}}=$$

$$=\sqrt{x^2-9}$$

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