Let f(x)=(2x^2-x). Give f '(1).

Question

anonymous · Answer

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

parthkohli · Answer

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

anonymous · Answer

1?

parthkohli · Answer

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

anonymous · Answer

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

anonymous · Answer

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.

parthkohli · Answer

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

anonymous · Answer

3

parthkohli · Answer

@cuzzin That includes differentiation too!

parthkohli · Answer

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

anonymous · Answer

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 :)

anonymous · Answer

why to make things more complicated

parthkohli · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

from differential.

anonymous · Answer

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

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