Differentiate $frac3x-6x$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
This should be fairly simple but I'm missing something. The way that I derived this equation was
$$f(x)=frac3x-6x$$
$$f'(x)=frac3(1x^0)-01x^0=frac31=3$$
What am I doing wrong?
derivatives
asked Aug 25 at 8:56
Pablo
33312
add a comment |Â
up vote
1
down vote
favorite
This should be fairly simple but I'm missing something. The way that I derived this equation was
$$f(x)=frac3x-6x$$
$$f'(x)=frac3(1x^0)-01x^0=frac31=3$$
What am I doing wrong?
derivatives
asked Aug 25 at 8:56
Pablo
33312
en.wikipedia.org/wiki/Quotient_rule
– Hans Lundmark
Aug 25 at 8:58
4
$$f(x) = x$$ is a function!
– Davide Morgante
Aug 25 at 9:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This should be fairly simple but I'm missing something. The way that I derived this equation was
$$f(x)=frac3x-6x$$
$$f'(x)=frac3(1x^0)-01x^0=frac31=3$$
What am I doing wrong?
derivatives
asked Aug 25 at 8:56
Pablo
33312
This should be fairly simple but I'm missing something. The way that I derived this equation was
$$f(x)=frac3x-6x$$
$$f'(x)=frac3(1x^0)-01x^0=frac31=3$$
What am I doing wrong?
derivatives
asked Aug 25 at 8:56
Pablo
33312
asked Aug 25 at 8:56
Pablo
33312
asked Aug 25 at 8:56
Pablo
33312
asked Aug 25 at 8:56
Pablo
33312
33312
en.wikipedia.org/wiki/Quotient_rule
– Hans Lundmark
Aug 25 at 8:58
4
$$f(x) = x$$ is a function!
– Davide Morgante
Aug 25 at 9:06
add a comment |Â
en.wikipedia.org/wiki/Quotient_rule
– Hans Lundmark
Aug 25 at 8:58
4
$$f(x) = x$$ is a function!
– Davide Morgante
Aug 25 at 9:06
en.wikipedia.org/wiki/Quotient_rule
– Hans Lundmark
Aug 25 at 8:58
en.wikipedia.org/wiki/Quotient_rule
– Hans Lundmark
Aug 25 at 8:58
4
4
$$f(x) = x$$ is a function!
– Davide Morgante
Aug 25 at 9:06
$$f(x) = x$$ is a function!
– Davide Morgante
Aug 25 at 9:06
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
By quotient rule
$$f’(x)=frac3x-(3x-6)x^2=frac 6x^2$$
as an alternative
$$f(x)=frac3x-6x=3-frac 6 ximplies f’(x)=frac 6x^2$$
answered Aug 25 at 9:05
gimusi
70k73786
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
add a comment |Â
up vote
2
down vote
You cannot differentiate numerator and denominator separately.
While you could use the quotient rule $(*)$, I usually try to avoid it when that's easily done:
$$left(frac3x-6xright)'=left(3-frac6xright)'=left(3-6x^-1right)'=6x^-2=frac6x^2$$With the quotient rule $(*)$:
$$left(frac3x-6xright)'=frac(3x-6)'x-(3x-6)x'x^2=frac3x-(3x-6)x^2=frac6x^2$$
$(*)$ the quotient rule tells you how to differentiate fractions:
$$left(fracf(x)g(x)right)'=fracf'(x)g(x)-f(x)g'(x)g(x)^2$$
Note how this is (very) different from $fracf'(x)g'(x)$...!
answered Aug 25 at 9:05
StackTD
20.5k1544
add a comment |Â
up vote
1
down vote
If one has to go by definition, then
$$f'(a) = lim _xto a left (frac3x-6x -frac3a-6aright ) cdot frac1x-a = lim _xto a frac6x-6aax(x-a) = frac6a^2 $$
answered Aug 25 at 9:05
Alvin Lepik
2,528921
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By quotient rule
$$f’(x)=frac3x-(3x-6)x^2=frac 6x^2$$
as an alternative
$$f(x)=frac3x-6x=3-frac 6 ximplies f’(x)=frac 6x^2$$
answered Aug 25 at 9:05
gimusi
70k73786
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
add a comment |Â
up vote
1
down vote
accepted
By quotient rule
$$f’(x)=frac3x-(3x-6)x^2=frac 6x^2$$
as an alternative
$$f(x)=frac3x-6x=3-frac 6 ximplies f’(x)=frac 6x^2$$
answered Aug 25 at 9:05
gimusi
70k73786
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By quotient rule
$$f’(x)=frac3x-(3x-6)x^2=frac 6x^2$$
as an alternative
$$f(x)=frac3x-6x=3-frac 6 ximplies f’(x)=frac 6x^2$$
answered Aug 25 at 9:05
gimusi
70k73786
By quotient rule
$$f’(x)=frac3x-(3x-6)x^2=frac 6x^2$$
as an alternative
$$f(x)=frac3x-6x=3-frac 6 ximplies f’(x)=frac 6x^2$$
answered Aug 25 at 9:05
gimusi
70k73786
answered Aug 25 at 9:05
gimusi
70k73786
answered Aug 25 at 9:05
gimusi
70k73786
answered Aug 25 at 9:05
gimusi
70k73786
70k73786
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
add a comment |Â
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify?
– Pablo
Aug 25 at 9:12
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
We simply have $$frac3x-6x=frac3xx-frac6x=3-frac6x$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$.
– gimusi
Aug 25 at 9:45
add a comment |Â
up vote
2
down vote
You cannot differentiate numerator and denominator separately.
While you could use the quotient rule $(*)$, I usually try to avoid it when that's easily done:
$$left(frac3x-6xright)'=left(3-frac6xright)'=left(3-6x^-1right)'=6x^-2=frac6x^2$$With the quotient rule $(*)$:
$$left(frac3x-6xright)'=frac(3x-6)'x-(3x-6)x'x^2=frac3x-(3x-6)x^2=frac6x^2$$
$(*)$ the quotient rule tells you how to differentiate fractions:
$$left(fracf(x)g(x)right)'=fracf'(x)g(x)-f(x)g'(x)g(x)^2$$
Note how this is (very) different from $fracf'(x)g'(x)$...!
answered Aug 25 at 9:05
StackTD
20.5k1544
add a comment |Â
up vote
2
down vote
You cannot differentiate numerator and denominator separately.
While you could use the quotient rule $(*)$, I usually try to avoid it when that's easily done:
$$left(frac3x-6xright)'=left(3-frac6xright)'=left(3-6x^-1right)'=6x^-2=frac6x^2$$With the quotient rule $(*)$:
$$left(frac3x-6xright)'=frac(3x-6)'x-(3x-6)x'x^2=frac3x-(3x-6)x^2=frac6x^2$$
$(*)$ the quotient rule tells you how to differentiate fractions:
$$left(fracf(x)g(x)right)'=fracf'(x)g(x)-f(x)g'(x)g(x)^2$$
Note how this is (very) different from $fracf'(x)g'(x)$...!
answered Aug 25 at 9:05
StackTD
20.5k1544
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You cannot differentiate numerator and denominator separately.
While you could use the quotient rule $(*)$, I usually try to avoid it when that's easily done:
$$left(frac3x-6xright)'=left(3-frac6xright)'=left(3-6x^-1right)'=6x^-2=frac6x^2$$With the quotient rule $(*)$:
$$left(frac3x-6xright)'=frac(3x-6)'x-(3x-6)x'x^2=frac3x-(3x-6)x^2=frac6x^2$$
$(*)$ the quotient rule tells you how to differentiate fractions:
$$left(fracf(x)g(x)right)'=fracf'(x)g(x)-f(x)g'(x)g(x)^2$$
Note how this is (very) different from $fracf'(x)g'(x)$...!
answered Aug 25 at 9:05
StackTD
20.5k1544
You cannot differentiate numerator and denominator separately.
While you could use the quotient rule $(*)$, I usually try to avoid it when that's easily done:
$$left(frac3x-6xright)'=left(3-frac6xright)'=left(3-6x^-1right)'=6x^-2=frac6x^2$$With the quotient rule $(*)$:
$$left(frac3x-6xright)'=frac(3x-6)'x-(3x-6)x'x^2=frac3x-(3x-6)x^2=frac6x^2$$
$(*)$ the quotient rule tells you how to differentiate fractions:
$$left(fracf(x)g(x)right)'=fracf'(x)g(x)-f(x)g'(x)g(x)^2$$
Note how this is (very) different from $fracf'(x)g'(x)$...!
answered Aug 25 at 9:05
StackTD
20.5k1544
edited Aug 25 at 9:11
answered Aug 25 at 9:05
StackTD
20.5k1544
answered Aug 25 at 9:05
StackTD
20.5k1544
answered Aug 25 at 9:05
StackTD
20.5k1544
20.5k1544
add a comment |Â
add a comment |Â
up vote
1
down vote
If one has to go by definition, then
$$f'(a) = lim _xto a left (frac3x-6x -frac3a-6aright ) cdot frac1x-a = lim _xto a frac6x-6aax(x-a) = frac6a^2 $$
answered Aug 25 at 9:05
Alvin Lepik
2,528921
add a comment |Â
up vote
1
down vote
If one has to go by definition, then
$$f'(a) = lim _xto a left (frac3x-6x -frac3a-6aright ) cdot frac1x-a = lim _xto a frac6x-6aax(x-a) = frac6a^2 $$
answered Aug 25 at 9:05
Alvin Lepik
2,528921
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If one has to go by definition, then
$$f'(a) = lim _xto a left (frac3x-6x -frac3a-6aright ) cdot frac1x-a = lim _xto a frac6x-6aax(x-a) = frac6a^2 $$
answered Aug 25 at 9:05
Alvin Lepik
2,528921
If one has to go by definition, then
$$f'(a) = lim _xto a left (frac3x-6x -frac3a-6aright ) cdot frac1x-a = lim _xto a frac6x-6aax(x-a) = frac6a^2 $$
answered Aug 25 at 9:05
Alvin Lepik
2,528921
answered Aug 25 at 9:05
Alvin Lepik
2,528921
answered Aug 25 at 9:05
Alvin Lepik
2,528921
answered Aug 25 at 9:05
Alvin Lepik
2,528921
2,528921
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2893913%2fdifferentiate-frac3x-6x%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
en.wikipedia.org/wiki/Quotient_rule
– Hans Lundmark
Aug 25 at 8:58
4
$$f(x) = x$$ is a function!
– Davide Morgante
Aug 25 at 9:06