derivatives equivalent somehow
$begingroup$
If I take the derivative of
$$frac{1}{1-x}$$
I get:
$$frac{1}{(1-x)^2}$$
If I take the derivative of
the same as $$frac{x}{1-x}$$
I also get
$$frac{1}{(1-x)^2}$$
Am I doing something wrong?
derivatives
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
$endgroup$
add a comment |
$begingroup$
If I take the derivative of
$$frac{1}{1-x}$$
I get:
$$frac{1}{(1-x)^2}$$
If I take the derivative of
the same as $$frac{x}{1-x}$$
I also get
$$frac{1}{(1-x)^2}$$
Am I doing something wrong?
derivatives
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
$endgroup$
add a comment |
$begingroup$
If I take the derivative of
$$frac{1}{1-x}$$
I get:
$$frac{1}{(1-x)^2}$$
If I take the derivative of
the same as $$frac{x}{1-x}$$
I also get
$$frac{1}{(1-x)^2}$$
Am I doing something wrong?
derivatives
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
$endgroup$
If I take the derivative of
$$frac{1}{1-x}$$
I get:
$$frac{1}{(1-x)^2}$$
If I take the derivative of
the same as $$frac{x}{1-x}$$
I also get
$$frac{1}{(1-x)^2}$$
Am I doing something wrong?
derivatives
derivatives
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
asked Jan 15 at 22:59
Jackson HartJackson Hart
5502727
5502727
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
Am I doing something wrong?
No, why? Indeed
$$
frac{x}{1-x}=frac{1-1+x}{1-x}=frac{1}{1-x}-1
$$
shows that the two functions differ by a constant, so they have the same derivative.
answered Jan 15 at 23:01
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
Well,
$$frac{1}{1-x}-frac{x}{1-x}=1$$
so
$$frac{d}{dx}frac{1}{1-x}-frac{d}{dx}frac{x}{1-x}=frac{d}{dx}1=0$$
answered Jan 15 at 23:02
SurbSurb
39.2k94479
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
$endgroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
answered Jan 15 at 23:02
DonAntonioDonAntonio
180k1495233
180k1495233
add a comment |
add a comment |
$begingroup$
Am I doing something wrong?
No, why? Indeed
$$
frac{x}{1-x}=frac{1-1+x}{1-x}=frac{1}{1-x}-1
$$
shows that the two functions differ by a constant, so they have the same derivative.
answered Jan 15 at 23:01
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
Am I doing something wrong?
No, why? Indeed
$$
frac{x}{1-x}=frac{1-1+x}{1-x}=frac{1}{1-x}-1
$$
shows that the two functions differ by a constant, so they have the same derivative.
answered Jan 15 at 23:01
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
Am I doing something wrong?
No, why? Indeed
$$
frac{x}{1-x}=frac{1-1+x}{1-x}=frac{1}{1-x}-1
$$
shows that the two functions differ by a constant, so they have the same derivative.
answered Jan 15 at 23:01
egregegreg
186k1486209
$endgroup$
Am I doing something wrong?
No, why? Indeed
$$
frac{x}{1-x}=frac{1-1+x}{1-x}=frac{1}{1-x}-1
$$
shows that the two functions differ by a constant, so they have the same derivative.
answered Jan 15 at 23:01
egregegreg
186k1486209
answered Jan 15 at 23:01
egregegreg
186k1486209
answered Jan 15 at 23:01
egregegreg
186k1486209
answered Jan 15 at 23:01
egregegreg
186k1486209
186k1486209
add a comment |
add a comment |
$begingroup$
Well,
$$frac{1}{1-x}-frac{x}{1-x}=1$$
so
$$frac{d}{dx}frac{1}{1-x}-frac{d}{dx}frac{x}{1-x}=frac{d}{dx}1=0$$
answered Jan 15 at 23:02
SurbSurb
39.2k94479
$endgroup$
add a comment |
$begingroup$
Well,
$$frac{1}{1-x}-frac{x}{1-x}=1$$
so
$$frac{d}{dx}frac{1}{1-x}-frac{d}{dx}frac{x}{1-x}=frac{d}{dx}1=0$$
answered Jan 15 at 23:02
SurbSurb
39.2k94479
$endgroup$
add a comment |
$begingroup$
Well,
$$frac{1}{1-x}-frac{x}{1-x}=1$$
so
$$frac{d}{dx}frac{1}{1-x}-frac{d}{dx}frac{x}{1-x}=frac{d}{dx}1=0$$
answered Jan 15 at 23:02
SurbSurb
39.2k94479
$endgroup$
Well,
$$frac{1}{1-x}-frac{x}{1-x}=1$$
so
$$frac{d}{dx}frac{1}{1-x}-frac{d}{dx}frac{x}{1-x}=frac{d}{dx}1=0$$
answered Jan 15 at 23:02
SurbSurb
39.2k94479
answered Jan 15 at 23:02
SurbSurb
39.2k94479
answered Jan 15 at 23:02
SurbSurb
39.2k94479
answered Jan 15 at 23:02
SurbSurb
39.2k94479
39.2k94479
add a comment |
add a comment |
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