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stipanrelix March 6th, 2018 08:25 AM

Trying to understand integration with very basic knowledge of derivation
 
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I have to show that this equation is true, I know I have to find the derivative of the right side, but I don't know how to find derivative of the square root.

Country Boy March 6th, 2018 11:50 AM

Do you know the power rule? The derivative of $x^n$ is $nx^{n-1}$.
$\sqrt{x}= x^{1/2}$ so the derivative is $\frac{1}{2}x^{-1/2}= \frac{1}{2\sqrt{x}}$.
Further, the derivative of ln(x) is $\frac{1}{x}$.

Using the "chain rule", the derivative of $ln\left(x+ \sqrt{x^2+ 1}\right)$ is
$\frac{1}{x+ \sqrt{x^2+ 1}}$ times the derivative of $x+ \sqrt{x^2+ 1}= x+ (x^2+ 1)^{1/2}$. And that derivative is $1+ (1/2)(x^2+1)^{-1/2}(2x)$$= 1+ \frac{x}{\sqrt{x^2+ 1}}$.

So the derivative you seek is
$\frac{1+ \frac{x}{\sqrt{x^2+ 1}}}{x+ \sqrt{x^2+ 1}}$

Multiplying numerator and denominator by $\sqrt{x^2+ 1}$ simplifies it slightly to $\frac{x+ \sqrt{x^2+ 1}}{x\sqrt{x^2+ 1}+ 1}$

greg1313 March 6th, 2018 12:41 PM

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


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