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Differntiation of ln 1/x

Hi can someone please help with this, I've got a different answer to what I'm supposed to.

y=ln1x=lnx1y= ln \frac{1}{x}= ln x^{-1}
dydx=(1x1)(1x2)\frac{dy}{dx} = (\frac{1}{x^{-1}})( -1x^{-2})
=1x2x1=\frac{-1x^{-2}}{x^{-1}}

The answer is supposed to be 11x-1\frac{1}{x} according to the book but I can't see how they got it.
y=ln(x1)=ln(x)y = ln(x^{-1}) = -ln(x)

start it off like that, its easier

also your answer is identical to the one they've given

1x2x1=11x21x1=11x\frac{-1x^{-2}}{x^{-1}} = -1 \frac{1}{x^{2}} \frac{1}{x^{-1}} = -1 \frac{1}{x}
(edited 13 years ago)
that's the same as what you got
Reply 3
Original post by mh1985
Hi can someone please help with this, I've got a different answer to what I'm supposed to.

y=ln1x=lnx1y= ln \frac{1}{x}= ln x^{-1}
dydx=(1x1)(1x2)\frac{dy}{dx} = (\frac{1}{x^{-1}})( -1x^{-2})
=1x2x1=\frac{-1x^{-2}}{x^{-1}}

The answer is supposed to be 11x-1\frac{1}{x} according to the book but I can't see how they got it.


Try simplifying your answer
Reply 4
Original post by mh1985
Hi can someone please help with this, I've got a different answer to what I'm supposed to.

y=ln1x=lnx1y= ln \frac{1}{x}= ln x^{-1}
dydx=(1x1)(1x2)\frac{dy}{dx} = (\frac{1}{x^{-1}})( -1x^{-2})
=1x2x1=\frac{-1x^{-2}}{x^{-1}}

The answer is supposed to be 11x-1\frac{1}{x} according to the book but I can't see how they got it.


x^-2 / x^-1 = x^-1
Cause its -2 - (-1) = -1

so your left with -1(x^-1)
= - 1/x
Reply 5
Original post by mh1985
Hi can someone please help with this, I've got a different answer to what I'm supposed to.





The answer is supposed to be according to the book but I can't see how they got it.


What you've done is correct, you just haven't simplified your answer fully. Try multiplying both the numerator and denominator by xx.

A faster way of approaching the problem would be to observe that ln(1x)=ln(x)ln(\frac{1}{x})=-ln(x), and going from there.

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