If f(x)= e^x, what is f′(e)?
I solved this and got
lim h->0 (e^(x+h) - e^e)/h
Can we have e^e because when I check in my calculator it shows as error!
I solved this and got
lim h->0 (e^(x+h) - e^e)/h
Can we have e^e because when I check in my calculator it shows as error!
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You can certainly have e^e. e is just a number, so e^e is just another number. I think you error may be that you still have an x in the derivative. It should be
f '(e) = Lim h-> 0 [e^(e + h) - e^(e)] / [h]
Of course, noticing that the derivative of e^x is still e^x, f '(e) = e^e, which will also be the value you obtain from evaluating the above limit.
Done!
f '(e) = Lim h-> 0 [e^(e + h) - e^(e)] / [h]
Of course, noticing that the derivative of e^x is still e^x, f '(e) = e^e, which will also be the value you obtain from evaluating the above limit.
Done!
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I have no idea why you have e^e in your expression. It should be e^x:
[e^(x+h) - e^x]/h = e^x [e^h - 1] / h
Now lim h->0 [e^h - 1]/h = 1, from which it follows that f'(x) = e^x, so f'(e) = e^e. This limit can be proven in a number of ways, though the easiest is probably to use the second definition of the exponential function on my reference.
[e^(x+h) - e^x]/h = e^x [e^h - 1] / h
Now lim h->0 [e^h - 1]/h = 1, from which it follows that f'(x) = e^x, so f'(e) = e^e. This limit can be proven in a number of ways, though the easiest is probably to use the second definition of the exponential function on my reference.
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f(x)= e^x
f '(x)=e^x also
f '(x)=e^x also