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
5162626
$endgroup$
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
5162626
$endgroup$
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
5162626
$endgroup$
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
5162626
asked Jan 15 at 22:59
Jackson HartJackson Hart
5162626
asked Jan 15 at 22:59
Jackson HartJackson Hart
5162626
asked Jan 15 at 22:59
Jackson HartJackson Hart
5162626
asked Jan 15 at 22:59
Jackson HartJackson Hart
5162626
5162626
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
178k1493229
$endgroup$
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
181k1485203
$endgroup$
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
37.7k94375
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
$endgroup$
Hint:
$$frac x{1-x}=-1+frac1{1-x}$$
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
answered Jan 15 at 23:02
DonAntonioDonAntonio
178k1493229
178k1493229
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
181k1485203
$endgroup$
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
181k1485203
$endgroup$
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
181k1485203
$endgroup$
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
181k1485203
answered Jan 15 at 23:01
egregegreg
181k1485203
answered Jan 15 at 23:01
egregegreg
181k1485203
answered Jan 15 at 23:01
egregegreg
181k1485203
181k1485203
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
37.7k94375
$endgroup$
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
37.7k94375
$endgroup$
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
37.7k94375
$endgroup$
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
37.7k94375
answered Jan 15 at 23:02
SurbSurb
37.7k94375
answered Jan 15 at 23:02
SurbSurb
37.7k94375
answered Jan 15 at 23:02
SurbSurb
37.7k94375
37.7k94375
add a comment |
add a comment |
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