cuzzin
Let f(x)=(2x^2-x). Give f '(1).
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cuzzin
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I'm a little stumped on this "Give f '(1) problems.
ParthKohli
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First, find the derivative. What is \(f'(x)\)?
cuzzin
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1?
ParthKohli
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No, no. It'd be\[f'(x) = {d \over dx}2x^2 - x\]Do you know what derivative is?
febylailani
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Let f(x)=(2x^2-x). Give f '(1).
f'(x) = 4x
f'(1) = 4(1) = 4 . am i right?
cuzzin
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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
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@febylailani I should point out that,\[f'(x) = 4x - 1\]
akash809
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3
ParthKohli
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@cuzzin That includes differentiation too!
ParthKohli
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\[f'(x) = 4x - 1\]\[f'(1) = 4(1) - 1\]
Omniscience
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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 :)
akash809
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why to make things more complicated
ParthKohli
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@Omniscience is pretty well till now, and then the power rule!:)
febylailani
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ah!! yup! sorry i forgot. careless -_- aaaaaccckkkkk
cuzzin
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Ok, where does the (4x-1) come from?
febylailani
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from differential.
febylailani
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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