Question

Find the derivative of the function f(x), below. It may be to your advantage to simplify first.

Answer 1

\[f(x)=\frac{ x^8 }{ 2^x }\]

Answer 2

why? use quotient rule i think would be a good idea

Answer 3

yes, I knew that fact that I had to use qutient rule but its 2^x.....

Answer 4

I have to also use chain rule?

Answer 5

I would rewrite it as:
\[f(x) = x^8 2^{-x}\]
as 1/2^x = 2^(-x)
Therefore you have a product rule giving:
\[(\frac{d}{dx}x^8)2^{-x} + x^8 \frac{d}{dx}2^{-x}\]
Now what is that :P
Useful:
\[\frac{d}{dx}a^x = \ln(a) a^x\]

Answer 6

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=x^8,f'(x)=8x^7, g(x)=2^x, g'(x)=2^x\ln(2)\]

Answer 7

Which you get by applying the change of base formula.

Answer 8

so do i use product or qutoinent.....?

Answer 9

or what @malevolence19 said, if you are allergic to the quotient rule

Answer 10

Tisk tisk satellite, assuming that f'g = g f' xP Sorry I'm in quantum and the fluttering operators...

Answer 11

i am partial to the quotient rule because when you get done you are usually in better shape to compute, for example, the critical points
otherwise if you use the product rule you then have the pesky task of adding up the fractions

Answer 12

Thank you for both of you!

Answer 13

\[\frac{ (x^8)(2^xln(2))-(2^x)(8x^7) }{ (2^x)^2 }\]
am i right?? @satellite73