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How do you differentiate #f(x)= ln(2x+1)^(-1/2) #?

Answer 1

Cameron Alexander

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

#d/dx sqrt(ln(2x+1)) = 1/(2sqrt(ln(2x+1)))*d/dxln(2x+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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Answer 2

Evelyn Agard

To differentiate ( f(x) = \ln((2x + 1)^{-1/2}) ), you can use the chain rule and the derivative of the natural logarithm function:

( f'(x) = \frac{d}{dx}[\ln((2x + 1)^{-1/2})] )

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

( = \frac{1}{(2x + 1)^{-1/2}} \cdot \frac{-1}{2(2x + 1)^{3/2}} \cdot \frac{d}{dx}(2x + 1) )

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

Therefore, ( f'(x) = \frac{-1}{2(2x + 1)} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.