# Thread: Finding the derivatives of functions

1. ## Finding the derivatives of functions

I have 2 problems I was confused with.
1. f(x)=(2x2-3x+1)/x
2. f(x)=x4/5

This is what I did for the first one:
I moved the x up, which turns into 2x2-3x+1+x-1
Then I multipled the coefficient by the exponents, and subtracted 1 from the exponent: 4x-3+1+x-2
But I was wondering, what do you do for the 1 here?

And for the second one:
Same thing, multipled exponent into coefficient: (4/5)x-1/5
Then 1/(4/5x5)
Then I got lost here... the answer in the book is: 4/(5x1/5)
It looks like the denominator just had the exponent switched into the positive, is that the right way to move it though?

2. ## Re: Finding the derivatives of functions

$\frac{2x^2 - 3x + 1}{x} = \frac{2x^2}{x} - \frac{3x}{x} + \frac{1}{x} = 2x - 3 + x^{-1}$ , now take the derivative.

$\frac{4}{5} x^{-\frac{1}{5}} = \frac{4}{5x^{\frac{1}{5}}}$ , only the x has a negative exponent

3. ## Re: Finding the derivatives of functions

Originally Posted by skeeter

$\frac{2x^2 - 3x + 1}{x} = \frac{2x^2}{x} - \frac{3x}{x} + \frac{1}{x} = 2x - 3 + x^{-1}$ , now take the derivative.

$\frac{4}{5} x^{-\frac{1}{5}} = \frac{4}{5x^{\frac{1}{5}}}$ , only the x has a negative exponent
Oh thanks!
So then the derivative would be 2-x-2, which then becomes 2-(1/x2) right?

4. ## Re: Finding the derivatives of functions

Originally Posted by Chaim
So then the derivative would be 2-x-2, which then becomes 2-(1/x2) right?
correct

5. ## Re: Finding the derivatives of functions

Originally Posted by skeeter
correct
Thanks!

Now I think I'm starting to get how to find derivatives of a function.