proof of product rule of derivatives using first principle?

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How can I prove the product rule of derivatives using the first principle?



$$fracd (f(x) g(x))d x
= left( fracd f(x)d x g(x) + fracd g(x)d x f(x) right)$$



Sorry if i used the wrong symbol for differential (I used delta), as I was unable to find the straight "d" on the web.




calculus differential




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  • 2




    Maybe you meant $fracddx(f(x)g(x))= fracdfdxg+ fracdgdxf$. If so you just simply use the ‘d’
    – MJW
    17 hours ago











  • yes, I meant that only. but i wasn't able to find the latex symbol for 'd'
    – Daksh Miglani
    17 hours ago










  • @MJW thanks, updated :))
    – Daksh Miglani
    17 hours ago






  • 1




    rm d gives $rm d$ but using $d$ is also acceptable.
    – Rócherz
    17 hours ago











  • @mvw yes, the limit definition of a derivative.
    – Daksh Miglani
    17 hours ago














up vote
0
down vote

favorite












How can I prove the product rule of derivatives using the first principle?



$$fracd (f(x) g(x))d x
= left( fracd f(x)d x g(x) + fracd g(x)d x f(x) right)$$



Sorry if i used the wrong symbol for differential (I used delta), as I was unable to find the straight "d" on the web.




calculus differential




share|cite|improve this question

















  • 2




    Maybe you meant $fracddx(f(x)g(x))= fracdfdxg+ fracdgdxf$. If so you just simply use the ‘d’
    – MJW
    17 hours ago











  • yes, I meant that only. but i wasn't able to find the latex symbol for 'd'
    – Daksh Miglani
    17 hours ago










  • @MJW thanks, updated :))
    – Daksh Miglani
    17 hours ago






  • 1




    rm d gives $rm d$ but using $d$ is also acceptable.
    – Rócherz
    17 hours ago











  • @mvw yes, the limit definition of a derivative.
    – Daksh Miglani
    17 hours ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











How can I prove the product rule of derivatives using the first principle?



$$fracd (f(x) g(x))d x
= left( fracd f(x)d x g(x) + fracd g(x)d x f(x) right)$$



Sorry if i used the wrong symbol for differential (I used delta), as I was unable to find the straight "d" on the web.




calculus differential




share|cite|improve this question













How can I prove the product rule of derivatives using the first principle?



$$fracd (f(x) g(x))d x
= left( fracd f(x)d x g(x) + fracd g(x)d x f(x) right)$$



Sorry if i used the wrong symbol for differential (I used delta), as I was unable to find the straight "d" on the web.





calculus differential





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 5 hours ago
























asked 17 hours ago









Daksh Miglani

1227




1227







  • 2




    Maybe you meant $fracddx(f(x)g(x))= fracdfdxg+ fracdgdxf$. If so you just simply use the ‘d’
    – MJW
    17 hours ago











  • yes, I meant that only. but i wasn't able to find the latex symbol for 'd'
    – Daksh Miglani
    17 hours ago










  • @MJW thanks, updated :))
    – Daksh Miglani
    17 hours ago






  • 1




    rm d gives $rm d$ but using $d$ is also acceptable.
    – Rócherz
    17 hours ago











  • @mvw yes, the limit definition of a derivative.
    – Daksh Miglani
    17 hours ago












  • 2




    Maybe you meant $fracddx(f(x)g(x))= fracdfdxg+ fracdgdxf$. If so you just simply use the ‘d’
    – MJW
    17 hours ago











  • yes, I meant that only. but i wasn't able to find the latex symbol for 'd'
    – Daksh Miglani
    17 hours ago










  • @MJW thanks, updated :))
    – Daksh Miglani
    17 hours ago






  • 1




    rm d gives $rm d$ but using $d$ is also acceptable.
    – Rócherz
    17 hours ago











  • @mvw yes, the limit definition of a derivative.
    – Daksh Miglani
    17 hours ago







2




2




Maybe you meant $fracddx(f(x)g(x))= fracdfdxg+ fracdgdxf$. If so you just simply use the ‘d’
– MJW
17 hours ago





Maybe you meant $fracddx(f(x)g(x))= fracdfdxg+ fracdgdxf$. If so you just simply use the ‘d’
– MJW
17 hours ago













yes, I meant that only. but i wasn't able to find the latex symbol for 'd'
– Daksh Miglani
17 hours ago




yes, I meant that only. but i wasn't able to find the latex symbol for 'd'
– Daksh Miglani
17 hours ago












@MJW thanks, updated :))
– Daksh Miglani
17 hours ago




@MJW thanks, updated :))
– Daksh Miglani
17 hours ago




1




1




rm d gives $rm d$ but using $d$ is also acceptable.
– Rócherz
17 hours ago





rm d gives $rm d$ but using $d$ is also acceptable.
– Rócherz
17 hours ago













@mvw yes, the limit definition of a derivative.
– Daksh Miglani
17 hours ago




@mvw yes, the limit definition of a derivative.
– Daksh Miglani
17 hours ago










1 Answer
1






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4
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There are two ways of stating the first principle.
The first one is $$fracrm df(x)rm dx =lim_hto 0 fracf(x+h)-f(x)h.$$
Then beginalign
fracrm dbig(f(x)g(x)big)rm dx
&=lim_hto 0 fracf(x+h)g(x+h) -f(x)g(x)h\
&=lim_hto 0 fracf(x+h)g(x+h)colorblue-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)h\
&=lim_hto 0 left( fracf(x+h)colorblue-f(x)h cdot g(x+h)
+f(x) cdot fraccolorblueg(x+h) -g(x)h right)\
&=lim_hto 0 fracf(x+h)colorblue-f(x)h cdot lim_hto 0 g(x+h)
+lim_hto 0 f(x) cdot lim_hto 0 fraccolorblueg(x+h)-g(x)h\
&=fracrm df(x)rm dx cdot g(x)
+f(x) cdot fracrm dg(x)rm dx.endalign



The second way is $$f'(a) =lim_xto a fracf(x)-f(a)x-a.$$
Then beginalign
(fg)'(a) &=lim_xto a fracf(x)g(x) -f(a)g(a)x-a\
&=lim_xto a fracf(x)g(x)colorblue-f(a)g(x)+f(a)g(x)-f(a)g(a)x-a\
&=lim_xto a left( fracf(x)colorblue-f(a)x-a cdot g(x)
+f(a) cdot fraccolorblueg(x)-g(a)x-a right)\
&=lim_xto a fracf(x)colorblue-f(a)x-a cdot lim_xto a g(x)
+lim_xto a f(a) cdot lim_xto a fraccolorblueg(x)-g(a)x-a\
&= f'(a) cdot g(a) +f(a) cdot g'(a).endalign






share|cite|improve this answer























  • i think there needs to be a limit as well?
    – Daksh Miglani
    16 hours ago










  • oh yeah, you edited. :)
    – Daksh Miglani
    16 hours ago






  • 3




    Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
    – Math_QED
    16 hours ago










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1 Answer
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1 Answer
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active

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active

oldest

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up vote
4
down vote



accepted










There are two ways of stating the first principle.
The first one is $$fracrm df(x)rm dx =lim_hto 0 fracf(x+h)-f(x)h.$$
Then beginalign
fracrm dbig(f(x)g(x)big)rm dx
&=lim_hto 0 fracf(x+h)g(x+h) -f(x)g(x)h\
&=lim_hto 0 fracf(x+h)g(x+h)colorblue-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)h\
&=lim_hto 0 left( fracf(x+h)colorblue-f(x)h cdot g(x+h)
+f(x) cdot fraccolorblueg(x+h) -g(x)h right)\
&=lim_hto 0 fracf(x+h)colorblue-f(x)h cdot lim_hto 0 g(x+h)
+lim_hto 0 f(x) cdot lim_hto 0 fraccolorblueg(x+h)-g(x)h\
&=fracrm df(x)rm dx cdot g(x)
+f(x) cdot fracrm dg(x)rm dx.endalign



The second way is $$f'(a) =lim_xto a fracf(x)-f(a)x-a.$$
Then beginalign
(fg)'(a) &=lim_xto a fracf(x)g(x) -f(a)g(a)x-a\
&=lim_xto a fracf(x)g(x)colorblue-f(a)g(x)+f(a)g(x)-f(a)g(a)x-a\
&=lim_xto a left( fracf(x)colorblue-f(a)x-a cdot g(x)
+f(a) cdot fraccolorblueg(x)-g(a)x-a right)\
&=lim_xto a fracf(x)colorblue-f(a)x-a cdot lim_xto a g(x)
+lim_xto a f(a) cdot lim_xto a fraccolorblueg(x)-g(a)x-a\
&= f'(a) cdot g(a) +f(a) cdot g'(a).endalign






share|cite|improve this answer























  • i think there needs to be a limit as well?
    – Daksh Miglani
    16 hours ago










  • oh yeah, you edited. :)
    – Daksh Miglani
    16 hours ago






  • 3




    Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
    – Math_QED
    16 hours ago














up vote
4
down vote



accepted










There are two ways of stating the first principle.
The first one is $$fracrm df(x)rm dx =lim_hto 0 fracf(x+h)-f(x)h.$$
Then beginalign
fracrm dbig(f(x)g(x)big)rm dx
&=lim_hto 0 fracf(x+h)g(x+h) -f(x)g(x)h\
&=lim_hto 0 fracf(x+h)g(x+h)colorblue-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)h\
&=lim_hto 0 left( fracf(x+h)colorblue-f(x)h cdot g(x+h)
+f(x) cdot fraccolorblueg(x+h) -g(x)h right)\
&=lim_hto 0 fracf(x+h)colorblue-f(x)h cdot lim_hto 0 g(x+h)
+lim_hto 0 f(x) cdot lim_hto 0 fraccolorblueg(x+h)-g(x)h\
&=fracrm df(x)rm dx cdot g(x)
+f(x) cdot fracrm dg(x)rm dx.endalign



The second way is $$f'(a) =lim_xto a fracf(x)-f(a)x-a.$$
Then beginalign
(fg)'(a) &=lim_xto a fracf(x)g(x) -f(a)g(a)x-a\
&=lim_xto a fracf(x)g(x)colorblue-f(a)g(x)+f(a)g(x)-f(a)g(a)x-a\
&=lim_xto a left( fracf(x)colorblue-f(a)x-a cdot g(x)
+f(a) cdot fraccolorblueg(x)-g(a)x-a right)\
&=lim_xto a fracf(x)colorblue-f(a)x-a cdot lim_xto a g(x)
+lim_xto a f(a) cdot lim_xto a fraccolorblueg(x)-g(a)x-a\
&= f'(a) cdot g(a) +f(a) cdot g'(a).endalign






share|cite|improve this answer























  • i think there needs to be a limit as well?
    – Daksh Miglani
    16 hours ago










  • oh yeah, you edited. :)
    – Daksh Miglani
    16 hours ago






  • 3




    Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
    – Math_QED
    16 hours ago












up vote
4
down vote



accepted







up vote
4
down vote



accepted






There are two ways of stating the first principle.
The first one is $$fracrm df(x)rm dx =lim_hto 0 fracf(x+h)-f(x)h.$$
Then beginalign
fracrm dbig(f(x)g(x)big)rm dx
&=lim_hto 0 fracf(x+h)g(x+h) -f(x)g(x)h\
&=lim_hto 0 fracf(x+h)g(x+h)colorblue-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)h\
&=lim_hto 0 left( fracf(x+h)colorblue-f(x)h cdot g(x+h)
+f(x) cdot fraccolorblueg(x+h) -g(x)h right)\
&=lim_hto 0 fracf(x+h)colorblue-f(x)h cdot lim_hto 0 g(x+h)
+lim_hto 0 f(x) cdot lim_hto 0 fraccolorblueg(x+h)-g(x)h\
&=fracrm df(x)rm dx cdot g(x)
+f(x) cdot fracrm dg(x)rm dx.endalign



The second way is $$f'(a) =lim_xto a fracf(x)-f(a)x-a.$$
Then beginalign
(fg)'(a) &=lim_xto a fracf(x)g(x) -f(a)g(a)x-a\
&=lim_xto a fracf(x)g(x)colorblue-f(a)g(x)+f(a)g(x)-f(a)g(a)x-a\
&=lim_xto a left( fracf(x)colorblue-f(a)x-a cdot g(x)
+f(a) cdot fraccolorblueg(x)-g(a)x-a right)\
&=lim_xto a fracf(x)colorblue-f(a)x-a cdot lim_xto a g(x)
+lim_xto a f(a) cdot lim_xto a fraccolorblueg(x)-g(a)x-a\
&= f'(a) cdot g(a) +f(a) cdot g'(a).endalign






share|cite|improve this answer















There are two ways of stating the first principle.
The first one is $$fracrm df(x)rm dx =lim_hto 0 fracf(x+h)-f(x)h.$$
Then beginalign
fracrm dbig(f(x)g(x)big)rm dx
&=lim_hto 0 fracf(x+h)g(x+h) -f(x)g(x)h\
&=lim_hto 0 fracf(x+h)g(x+h)colorblue-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)h\
&=lim_hto 0 left( fracf(x+h)colorblue-f(x)h cdot g(x+h)
+f(x) cdot fraccolorblueg(x+h) -g(x)h right)\
&=lim_hto 0 fracf(x+h)colorblue-f(x)h cdot lim_hto 0 g(x+h)
+lim_hto 0 f(x) cdot lim_hto 0 fraccolorblueg(x+h)-g(x)h\
&=fracrm df(x)rm dx cdot g(x)
+f(x) cdot fracrm dg(x)rm dx.endalign



The second way is $$f'(a) =lim_xto a fracf(x)-f(a)x-a.$$
Then beginalign
(fg)'(a) &=lim_xto a fracf(x)g(x) -f(a)g(a)x-a\
&=lim_xto a fracf(x)g(x)colorblue-f(a)g(x)+f(a)g(x)-f(a)g(a)x-a\
&=lim_xto a left( fracf(x)colorblue-f(a)x-a cdot g(x)
+f(a) cdot fraccolorblueg(x)-g(a)x-a right)\
&=lim_xto a fracf(x)colorblue-f(a)x-a cdot lim_xto a g(x)
+lim_xto a f(a) cdot lim_xto a fraccolorblueg(x)-g(a)x-a\
&= f'(a) cdot g(a) +f(a) cdot g'(a).endalign







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited 10 hours ago


























answered 16 hours ago









Rócherz

2,1811417




2,1811417











  • i think there needs to be a limit as well?
    – Daksh Miglani
    16 hours ago










  • oh yeah, you edited. :)
    – Daksh Miglani
    16 hours ago






  • 3




    Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
    – Math_QED
    16 hours ago
















  • i think there needs to be a limit as well?
    – Daksh Miglani
    16 hours ago










  • oh yeah, you edited. :)
    – Daksh Miglani
    16 hours ago






  • 3




    Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
    – Math_QED
    16 hours ago















i think there needs to be a limit as well?
– Daksh Miglani
16 hours ago




i think there needs to be a limit as well?
– Daksh Miglani
16 hours ago












oh yeah, you edited. :)
– Daksh Miglani
16 hours ago




oh yeah, you edited. :)
– Daksh Miglani
16 hours ago




3




3




Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
– Math_QED
16 hours ago




Note that continuity in $x$ was used to go to the last line, which is a consequence of differentiability in $x$.
– Math_QED
16 hours ago












 

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