Derivative of a logarithmic function with $frac{C}{x}$












0












$begingroup$


The function is



$$G(x)=4^{frac{C}{x}}$$
I have $u=frac{C}{x}$, then I calculate $$frac{d}{du}4^u=u(4)^{u-1}$$
But does $$frac{d}{dx}frac{C}{x}=C$$
because I don't know if $C$ is a constant or a independent variable.






calculus logarithms







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$endgroup$












  • $begingroup$
    $frac{d}{du}4^u=4^uln4ne u(4)^{u-1}$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 12:01


















0












$begingroup$


The function is



$$G(x)=4^{frac{C}{x}}$$
I have $u=frac{C}{x}$, then I calculate $$frac{d}{du}4^u=u(4)^{u-1}$$
But does $$frac{d}{dx}frac{C}{x}=C$$
because I don't know if $C$ is a constant or a independent variable.






calculus logarithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    $frac{d}{du}4^u=4^uln4ne u(4)^{u-1}$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 12:01
















0












0








0





$begingroup$


The function is



$$G(x)=4^{frac{C}{x}}$$
I have $u=frac{C}{x}$, then I calculate $$frac{d}{du}4^u=u(4)^{u-1}$$
But does $$frac{d}{dx}frac{C}{x}=C$$
because I don't know if $C$ is a constant or a independent variable.






calculus logarithms







share|cite|improve this question









$endgroup$




The function is



$$G(x)=4^{frac{C}{x}}$$
I have $u=frac{C}{x}$, then I calculate $$frac{d}{du}4^u=u(4)^{u-1}$$
But does $$frac{d}{dx}frac{C}{x}=C$$
because I don't know if $C$ is a constant or a independent variable.






calculus logarithms




calculus logarithms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 11:27









Eric BrownEric Brown

757




757












  • $begingroup$
    $frac{d}{du}4^u=4^uln4ne u(4)^{u-1}$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 12:01




















  • $begingroup$
    $frac{d}{du}4^u=4^uln4ne u(4)^{u-1}$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 12:01


















$begingroup$
$frac{d}{du}4^u=4^uln4ne u(4)^{u-1}$
$endgroup$
– Shubham Johri
Dec 11 '18 at 12:01






$begingroup$
$frac{d}{du}4^u=4^uln4ne u(4)^{u-1}$
$endgroup$
– Shubham Johri
Dec 11 '18 at 12:01












1 Answer
1






active

oldest

votes


















2












$begingroup$

HINT



if $f(x) = a^x$ you can write



$$
f(x) = e^{ln a^x} = e^{x ln a}
$$



So the derivative is



$$
frac{{rm d}f}{{rm d}x} = (ln a )e^{xln a} = a^x ln a
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you use the terms that are in my question cause I'm confused about what I'm reading.
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 11:32










  • $begingroup$
    @EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
    $endgroup$
    – caverac
    Dec 11 '18 at 11:37












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

HINT



if $f(x) = a^x$ you can write



$$
f(x) = e^{ln a^x} = e^{x ln a}
$$



So the derivative is



$$
frac{{rm d}f}{{rm d}x} = (ln a )e^{xln a} = a^x ln a
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you use the terms that are in my question cause I'm confused about what I'm reading.
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 11:32










  • $begingroup$
    @EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
    $endgroup$
    – caverac
    Dec 11 '18 at 11:37
















2












$begingroup$

HINT



if $f(x) = a^x$ you can write



$$
f(x) = e^{ln a^x} = e^{x ln a}
$$



So the derivative is



$$
frac{{rm d}f}{{rm d}x} = (ln a )e^{xln a} = a^x ln a
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you use the terms that are in my question cause I'm confused about what I'm reading.
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 11:32










  • $begingroup$
    @EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
    $endgroup$
    – caverac
    Dec 11 '18 at 11:37














2












2








2





$begingroup$

HINT



if $f(x) = a^x$ you can write



$$
f(x) = e^{ln a^x} = e^{x ln a}
$$



So the derivative is



$$
frac{{rm d}f}{{rm d}x} = (ln a )e^{xln a} = a^x ln a
$$






share|cite|improve this answer









$endgroup$



HINT



if $f(x) = a^x$ you can write



$$
f(x) = e^{ln a^x} = e^{x ln a}
$$



So the derivative is



$$
frac{{rm d}f}{{rm d}x} = (ln a )e^{xln a} = a^x ln a
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 11:30









caveraccaverac

14.8k31130




14.8k31130












  • $begingroup$
    Could you use the terms that are in my question cause I'm confused about what I'm reading.
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 11:32










  • $begingroup$
    @EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
    $endgroup$
    – caverac
    Dec 11 '18 at 11:37


















  • $begingroup$
    Could you use the terms that are in my question cause I'm confused about what I'm reading.
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 11:32










  • $begingroup$
    @EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
    $endgroup$
    – caverac
    Dec 11 '18 at 11:37
















$begingroup$
Could you use the terms that are in my question cause I'm confused about what I'm reading.
$endgroup$
– Eric Brown
Dec 11 '18 at 11:32




$begingroup$
Could you use the terms that are in my question cause I'm confused about what I'm reading.
$endgroup$
– Eric Brown
Dec 11 '18 at 11:32












$begingroup$
@EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
$endgroup$
– caverac
Dec 11 '18 at 11:37




$begingroup$
@EricBrown $$ frac{{rm d}}{{rm d}x} 4^{C/x} = frac{{rm d}}{{rm d}x}e^{Cln 4/x} = ... $$
$endgroup$
– caverac
Dec 11 '18 at 11:37


















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