## cuzzin 3 years ago Let f(x)=(2x^2-x). Give f '(1).

1. cuzzin

I'm a little stumped on this "Give f '(1) problems.

2. ParthKohli

First, find the derivative. What is $$f'(x)$$?

3. cuzzin

1?

4. ParthKohli

No, no. It'd be$f'(x) = {d \over dx}2x^2 - x$Do you know what derivative is?

5. febylailani

Let f(x)=(2x^2-x). Give f '(1). f'(x) = 4x f'(1) = 4(1) = 4 . am i right?

6. cuzzin

Well we are just starting the chapter on derivatives, so I'm still figuring them out. Mostly we've been dealing with finding the slope of tangent and normal lines.

7. ParthKohli

@febylailani I should point out that,$f'(x) = 4x - 1$

8. akash809

3

9. ParthKohli

@cuzzin That includes differentiation too!

10. ParthKohli

$f'(x) = 4x - 1$$f'(1) = 4(1) - 1$

11. Omniscience

I think you should write it as: $\frac{d}{dx} \left(2x^{2}-x\right) or \frac{d}{dx} 2x^2-\frac{d}{dx} x$ Or people would get confused,which was evident :)

12. akash809

why to make things more complicated

13. ParthKohli

@Omniscience is pretty well till now, and then the power rule!:)

14. febylailani

ah!! yup! sorry i forgot. careless -_- aaaaaccckkkkk

15. cuzzin

Ok, where does the (4x-1) come from?

16. febylailani

from differential.

17. febylailani

but differential is the short way. before that, there's limit. the formula is a.n.x^(n-1). i.e. f(x) = 2x^3 ; a=2 ; n = 3 so, f'(x) = 2.3.x^(3-1) = 6x^2 cmiiw