How do you differentiate $f (x) = {ln (2 x + 1)}^{- \frac{1}{2}}$ $f(x)= ln(2x+1)^(-1/2)$ ?

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How do you differentiate $f (x) = {ln (2 x + 1)}^{- \frac{1}{2}}$ $f(x)= ln(2x+1)^(-1/2)$ ?

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Answer 1 · 2017-11-16T00:44:00

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

Explanation:

$\frac{d}{d x} \sqrt{ln (2 x + 1)} = \frac{1}{2 \sqrt{ln (2 x + 1)}} \cdot \frac{d}{d x} ln (2 x + 1)$ $d/dx sqrt(ln(2x+1)) = 1/(2sqrt(ln(2x+1)))*d/dxln(2x+1)$

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

$=1/(cancel(2)sqrt(ln(2x+1)))*1/(2x+1) *cancel(2)$

$=1/((2x+1)sqrt(ln(2x+1)))$

Basically, it is the chain rule a bunch of times.

I hope that helps!

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How do you differentiate $f (x) = {ln (2 x + 1)}^{- \frac{1}{2}}$ $f(x)= ln(2x+1)^(-1/2)$ ?

1 Answer

Explanation:

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How do you differentiate $f (x) = {ln (2 x + 1)}^{- \frac{1}{2}}$ $f(x)= ln(2x+1)^(-1/2)$ ?