I don't want the answer i want a method 
Hint: You can write 2^x as e^ln(2^x).
y = 2^x
1) first take logs so that
lny = ln(2^x)
2) apply log rules gives
lny =xln2
3) then differentiate both sides respect to x to give
(1/y)(dy/dx) = ln2
therefore
dy/dx = yln2
we already know that y = 2^x therefore
=(2^x)ln2
1) first take logs so that
lny = ln(2^x)
2) apply log rules gives
lny =xln2
3) then differentiate both sides respect to x to give
(1/y)(dy/dx) = ln2
therefore
dy/dx = yln2
we already know that y = 2^x therefore
=(2^x)ln2
(edited 11 years ago)
Since y = a ^ x (where a is any number)
ln y = ln(a ^ x) (ln both sides)
ln y = x ln a (use law of the logs on right hand side)
(1/y) * (dy/dx) = ln a (using implicit diffrentiation with respect to x)
dy/dx = y ln a (bring the y over)
dy/dx = (a ^ x) * ln a (y = a ^ x, so replace the y with a ^ x)
In this case, dy/dx = (ln 2) * 2 ^ x
I'm not sure how far you've got in maths, and whether you know implicit diffrentiation or how to diffrentiate ln yet, but I hope this helps, and reply if you want me to explain further...
ln y = ln(a ^ x) (ln both sides)
ln y = x ln a (use law of the logs on right hand side)
(1/y) * (dy/dx) = ln a (using implicit diffrentiation with respect to x)
dy/dx = y ln a (bring the y over)
dy/dx = (a ^ x) * ln a (y = a ^ x, so replace the y with a ^ x)
In this case, dy/dx = (ln 2) * 2 ^ x
I'm not sure how far you've got in maths, and whether you know implicit diffrentiation or how to diffrentiate ln yet, but I hope this helps, and reply if you want me to explain further...
(edited 11 years ago)
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