On differentiating $F(x)=ln(2x)$












2












$begingroup$


If we differentiate $F(x)=ln(2x)$ we will get $F'(x) =dfrac2{2x}$ after the shortcut $F'(x)=dfrac1x$, right?



Now if we integrate $F'(x)$ we will get $F(x)=ln(x)$ but also $F(x)=ln(2x)$. This means $ln(2x)=ln(x)$. What happened?






calculus derivatives







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$endgroup$








  • 2




    $begingroup$
    Antiderivatives are same upto constants.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 6:20


















2












$begingroup$


If we differentiate $F(x)=ln(2x)$ we will get $F'(x) =dfrac2{2x}$ after the shortcut $F'(x)=dfrac1x$, right?



Now if we integrate $F'(x)$ we will get $F(x)=ln(x)$ but also $F(x)=ln(2x)$. This means $ln(2x)=ln(x)$. What happened?






calculus derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Antiderivatives are same upto constants.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 6:20
















2












2








2


1



$begingroup$


If we differentiate $F(x)=ln(2x)$ we will get $F'(x) =dfrac2{2x}$ after the shortcut $F'(x)=dfrac1x$, right?



Now if we integrate $F'(x)$ we will get $F(x)=ln(x)$ but also $F(x)=ln(2x)$. This means $ln(2x)=ln(x)$. What happened?






calculus derivatives







share|cite|improve this question











$endgroup$




If we differentiate $F(x)=ln(2x)$ we will get $F'(x) =dfrac2{2x}$ after the shortcut $F'(x)=dfrac1x$, right?



Now if we integrate $F'(x)$ we will get $F(x)=ln(x)$ but also $F(x)=ln(2x)$. This means $ln(2x)=ln(x)$. What happened?






calculus derivatives




calculus derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 6:34









Saad

19.7k92352




19.7k92352










asked Dec 20 '18 at 6:11









Osaid Ibrahim Bani SalmanOsaid Ibrahim Bani Salman

244




244








  • 2




    $begingroup$
    Antiderivatives are same upto constants.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 6:20
















  • 2




    $begingroup$
    Antiderivatives are same upto constants.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 6:20










2




2




$begingroup$
Antiderivatives are same upto constants.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 6:20






$begingroup$
Antiderivatives are same upto constants.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 6:20












3 Answers
3






active

oldest

votes


















3












$begingroup$

When you integrate, there is an arbitrary constant $C$ added. So you only get
$$ln(2x)=ln(x)+C.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this is exactly right Bjorn, thanks.
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:25



















3












$begingroup$

But$$F(x)=ln(2x) =ln 2 + ln x$$
So when you integrate back you get $$ln x + C $$
where the constant $C$ should be the $ln 2$ term.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is exactly right, Ahmad Thanks. With recpect
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:26










  • $begingroup$
    im glad it was helpful @OsaidIbrahimBaniSalman
    $endgroup$
    – Ahmad Bazzi
    Dec 20 '18 at 6:43



















2












$begingroup$

The antiderivatives of $1/x$ are not uniquely determined ! They are given by



$$ln x +C.$$



Forthermore $ ln (2x)= ln 2 + ln x.$



Can you proceed ?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is true. Yes i can. Thanks
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:28










  • $begingroup$
    On a connected domain at least.
    $endgroup$
    – Math_QED
    Dec 20 '18 at 7:00











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

When you integrate, there is an arbitrary constant $C$ added. So you only get
$$ln(2x)=ln(x)+C.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this is exactly right Bjorn, thanks.
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:25
















3












$begingroup$

When you integrate, there is an arbitrary constant $C$ added. So you only get
$$ln(2x)=ln(x)+C.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this is exactly right Bjorn, thanks.
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:25














3












3








3





$begingroup$

When you integrate, there is an arbitrary constant $C$ added. So you only get
$$ln(2x)=ln(x)+C.$$






share|cite|improve this answer









$endgroup$



When you integrate, there is an arbitrary constant $C$ added. So you only get
$$ln(2x)=ln(x)+C.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 6:13









Bjørn Kjos-HanssenBjørn Kjos-Hanssen

2,097918




2,097918












  • $begingroup$
    this is exactly right Bjorn, thanks.
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:25


















  • $begingroup$
    this is exactly right Bjorn, thanks.
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:25
















$begingroup$
this is exactly right Bjorn, thanks.
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 6:25




$begingroup$
this is exactly right Bjorn, thanks.
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 6:25











3












$begingroup$

But$$F(x)=ln(2x) =ln 2 + ln x$$
So when you integrate back you get $$ln x + C $$
where the constant $C$ should be the $ln 2$ term.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is exactly right, Ahmad Thanks. With recpect
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:26










  • $begingroup$
    im glad it was helpful @OsaidIbrahimBaniSalman
    $endgroup$
    – Ahmad Bazzi
    Dec 20 '18 at 6:43
















3












$begingroup$

But$$F(x)=ln(2x) =ln 2 + ln x$$
So when you integrate back you get $$ln x + C $$
where the constant $C$ should be the $ln 2$ term.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is exactly right, Ahmad Thanks. With recpect
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:26










  • $begingroup$
    im glad it was helpful @OsaidIbrahimBaniSalman
    $endgroup$
    – Ahmad Bazzi
    Dec 20 '18 at 6:43














3












3








3





$begingroup$

But$$F(x)=ln(2x) =ln 2 + ln x$$
So when you integrate back you get $$ln x + C $$
where the constant $C$ should be the $ln 2$ term.






share|cite|improve this answer









$endgroup$



But$$F(x)=ln(2x) =ln 2 + ln x$$
So when you integrate back you get $$ln x + C $$
where the constant $C$ should be the $ln 2$ term.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 6:15









Ahmad BazziAhmad Bazzi

8,3622824




8,3622824








  • 1




    $begingroup$
    This is exactly right, Ahmad Thanks. With recpect
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:26










  • $begingroup$
    im glad it was helpful @OsaidIbrahimBaniSalman
    $endgroup$
    – Ahmad Bazzi
    Dec 20 '18 at 6:43














  • 1




    $begingroup$
    This is exactly right, Ahmad Thanks. With recpect
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:26










  • $begingroup$
    im glad it was helpful @OsaidIbrahimBaniSalman
    $endgroup$
    – Ahmad Bazzi
    Dec 20 '18 at 6:43








1




1




$begingroup$
This is exactly right, Ahmad Thanks. With recpect
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 6:26




$begingroup$
This is exactly right, Ahmad Thanks. With recpect
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 6:26












$begingroup$
im glad it was helpful @OsaidIbrahimBaniSalman
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 6:43




$begingroup$
im glad it was helpful @OsaidIbrahimBaniSalman
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 6:43











2












$begingroup$

The antiderivatives of $1/x$ are not uniquely determined ! They are given by



$$ln x +C.$$



Forthermore $ ln (2x)= ln 2 + ln x.$



Can you proceed ?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is true. Yes i can. Thanks
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:28










  • $begingroup$
    On a connected domain at least.
    $endgroup$
    – Math_QED
    Dec 20 '18 at 7:00
















2












$begingroup$

The antiderivatives of $1/x$ are not uniquely determined ! They are given by



$$ln x +C.$$



Forthermore $ ln (2x)= ln 2 + ln x.$



Can you proceed ?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is true. Yes i can. Thanks
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:28










  • $begingroup$
    On a connected domain at least.
    $endgroup$
    – Math_QED
    Dec 20 '18 at 7:00














2












2








2





$begingroup$

The antiderivatives of $1/x$ are not uniquely determined ! They are given by



$$ln x +C.$$



Forthermore $ ln (2x)= ln 2 + ln x.$



Can you proceed ?






share|cite|improve this answer









$endgroup$



The antiderivatives of $1/x$ are not uniquely determined ! They are given by



$$ln x +C.$$



Forthermore $ ln (2x)= ln 2 + ln x.$



Can you proceed ?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 6:18









FredFred

47.2k1849




47.2k1849












  • $begingroup$
    This is true. Yes i can. Thanks
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:28










  • $begingroup$
    On a connected domain at least.
    $endgroup$
    – Math_QED
    Dec 20 '18 at 7:00


















  • $begingroup$
    This is true. Yes i can. Thanks
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 6:28










  • $begingroup$
    On a connected domain at least.
    $endgroup$
    – Math_QED
    Dec 20 '18 at 7:00
















$begingroup$
This is true. Yes i can. Thanks
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 6:28




$begingroup$
This is true. Yes i can. Thanks
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 6:28












$begingroup$
On a connected domain at least.
$endgroup$
– Math_QED
Dec 20 '18 at 7:00




$begingroup$
On a connected domain at least.
$endgroup$
– Math_QED
Dec 20 '18 at 7:00


















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