How do you find the derivative of $- ln (x - {(x^{2} + 1)}^{\frac{1}{2}})$ $-ln(x-(x^2+1)^(1/2))$ ?

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How do you find the derivative of $- ln (x - {(x^{2} + 1)}^{\frac{1}{2}})$ $-ln(x-(x^2+1)^(1/2))$ ?

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Answer 1 · 2017-02-12T12:32:45

The derivative of $- ln (x - {(x^{2} + 1)}^{\frac{1}{2}})$ $-ln(x - (x^2+1)^(1/2))$ with respect to x is $\frac{1}{\sqrt{x^{2} + 1}} .$ $1/sqrt(x^2 +1).$ To solve this use the chain rule carefully. See the explanation for details.

Explanation:

Starting with:

$\frac{d}{d x} (- ln (x - {(x^{2} + 1)}^{\frac{1}{2}}))$ $d / dx (-ln(x - (x^2+1)^(1/2)))$

use the chain rule to get:

$\frac{d}{d x} (- ln (x - {(x^{2} + 1)}^{\frac{1}{2}}))$ $d / dx (-ln(x - (x^2+1)^(1/2)))$

$= - \frac{1}{x - {(x^{2} + 1)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x - {(x^{2} + 1)}^{\frac{1}{2}})$ $= -1/(x-(x^2+1)^(1/2)) *d/dx (x - (x^2+1)^(1/2))$

use the chain rule once again on the remaining derivative:

$= - \frac{1}{x - {(x^{2} + 1)}^{\frac{1}{2}}} \cdot (1 - \frac{1}{2} {(x^{2} + 1)}^{- \frac{1}{2}} (2 x))$ $= -1/(x-(x^2+1)^(1/2)) * (1 - 1/2(x^2+1)^(-1/2)(2x))$

Simplify:

$= \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - 1}{x - {(x^{2} + 1)}^{\frac{1}{2}}}$ $= (x/(x^2+1)^(1/2) -1 )/(x-(x^2+1)^(1/2))$

Note that $1 = \frac{{(x^{2} + 1)}^{\frac{1}{2}}}{{(x^{2} + 1)}^{\frac{1}{2}}}$ $1=(x^2+1)^(1/2) / (x^2+1)^(1/2)$ , then substitute this for 1:

$= \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{{(x^{2} + 1)}^{\frac{1}{2}}}{{(x^{2} + 1)}^{\frac{1}{2}}}}{x - {(x^{2} + 1)}^{\frac{1}{2}}}$ $= (x/(x^2+1)^(1/2) - (x^2+1)^(1/2) / (x^2+1)^(1/2) )/(x-(x^2+1)^(1/2))$

$= \frac{\frac{x - {(x^{2} + 1)}^{\frac{1}{2}}}{{(x^{2} + 1)}^{\frac{1}{2}}}}{x - {(x^{2} + 1)}^{\frac{1}{2}}}$ $= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) )/(x-(x^2+1)^(1/2))$

$= ⎛ ⎜ ⎝ \frac{x - {(x^{2} + 1)}^{\frac{1}{2}}}{{(x^{2} + 1)}^{\frac{1}{2}}} ⎞ ⎟ ⎠ \div (x - {(x^{2} + 1)}^{\frac{1}{2}})$ $= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) ) divide (x-(x^2+1)^(1/2))$

$= ⎛ ⎜ ⎝ \frac{x - {(x^{2} + 1)}^{\frac{1}{2}}}{{(x^{2} + 1)}^{\frac{1}{2}}} ⎞ ⎟ ⎠ \cdot \frac{1}{x - {(x^{2} + 1)}^{\frac{1}{2}}}$ $= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) ) * 1/(x-(x^2+1)^(1/2))$

$= \frac{x - {(x^{2} + 1)}^{\frac{1}{2}}}{x - {(x^{2} + 1)}^{\frac{1}{2}}} \cdot \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ $= (x - (x^2+1)^(1/2)) / (x-(x^2+1)^(1/2) ) * 1/((x^2+1)^(1/2))$

$= 1 \cdot \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ $= 1 * 1/((x^2+1)^(1/2))$

$= \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} = \frac{1}{\sqrt{(x^{2} + 1)}} .$ $= 1/((x^2+1)^(1/2)) = 1/sqrt((x^2+1)).$

Finally:

$\frac{d}{d x} (- ln (x - {(x^{2} + 1)}^{\frac{1}{2}})) = \frac{1}{\sqrt{(x^{2} + 1)}} .$ $d / dx (-ln(x - (x^2+1)^(1/2))) = 1/sqrt((x^2+1)).$

If you have any questions about the use of the chain rule or any other part of this solution, then please ask.

Rory.

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