User:Freezertennis3423

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Derivative of ln(x\sqrt{x^2+1}) :

Using chain rule... Derivative of x^2+1 \, is 2x \,, therefore derivative of \sqrt {x^2+1} is 2x {1 \over 2\sqrt {x^2+1}} , which simplifies to {2x \over 2\sqrt {x^2+1}} , which simplifies to {x \over \sqrt {x^2+1}}

Then using product rule: The derivative of x\sqrt{x^2+1} is \sqrt{x^2+1}+x{x \over \sqrt {x^2+1}} , which simplifies to \sqrt{x^2+1}+{x^2 \over \sqrt {x^2+1}} , which can be rewritten as {2x^2+1 \over \sqrt {x^2+1}} by finding a common denominator

Finally, using the chain rule, the derivative of ln(x\sqrt{x^2+1}) \, is {2x^2+1 \over \sqrt {x^2+1}} \times {1 \over {x\sqrt{x^2+1}}} , which simplifies to {2x^2+1} \over {x\sqrt{x^2+1}\sqrt{x^2+1}} or {2x^2+1} \over {x(x^2+1)} or {2x^2+1} \over {x^3+x} , which can be simplified to {x \over {x^2+1}}+{1 \over x} by using algebra