Is that (8^2) * x ? (which is what you have written.)
EZ, => 8^2
OR do you mean
8^(2 * x) ? (which is not what you have written)
>What is the derivative of 8^2x?
What you wrote means (8^2)*x or 64x. Did you mean 8^(2x)? If you are
taking calculus you really should be using parentheses when needed.
Guessing at what you mean my hint is to look in your calculus book for
the derivative of a^x and remember to use the chain rule.
--Lynn
>What is the derivative of 8^2x?
>
8^(2 x) = e^[(ln 64) x]
--
Stephen J. Herschkorn sjher...@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan
Wrong.
64.
Or perhaps you meant 8^(2x), in which
case rewrite this as exp(somnething) and
differentiate that.
>"Stephen J. Herschkorn" <sjher...@netscape.net> wrote in message
>news:43C6A409...@netscape.net...
>
>
>>Invalid...@gmail.com wrote:
>>
>>
>>
>>>What is the derivative of 8^2x?
>>>
>>>
>>>
>>8^(2 x) = e^[(ln 64) x]
>>
>Wrong.
>
Huh? What is wrong in that equality? (Note that, intentionally, I did
not answer the original question.)
You are right, my dyslexia
>>> What is the derivative of 8^2x?
Stephen J. Herschkorn wrote:
>> 8^(2 x) = e^[(ln 64) x]
Maulana Gott wrote:
> Wrong.
How so? Stephen J. Herschkorn's function isn't
equal to (8^2)x and it's not the derivative of
either (8^2)x nor the derivative of 8^(2x), but
the equation he wrote is true unless I'm missing
something in front of my nose (like I did a couple
of days ago). (If Robert Israel is reading, I did
learn that in l'Hopital's rule the assumption that
g(x) is not zero in a neighborhood of x=a follows
from the other assumptions, but I don't have time
now to post the details.)
Dave L. Renfro