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cuzzin
 2 years ago
Best ResponseYou've already chosen the best response.0I'm a little stumped on this "Give f '(1) problems.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0First, find the derivative. What is \(f'(x)\)?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0No, no. It'd be\[f'(x) = {d \over dx}2x^2  x\]Do you know what derivative is?

febylailani
 2 years ago
Best ResponseYou've already chosen the best response.0Let f(x)=(2x^2x). Give f '(1). f'(x) = 4x f'(1) = 4(1) = 4 . am i right?

cuzzin
 2 years ago
Best ResponseYou've already chosen the best response.0Well 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
 2 years ago
Best ResponseYou've already chosen the best response.0@febylailani I should point out that,\[f'(x) = 4x  1\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0@cuzzin That includes differentiation too!

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = 4x  1\]\[f'(1) = 4(1)  1\]

Omniscience
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.0why to make things more complicated

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0@Omniscience is pretty well till now, and then the power rule!:)

febylailani
 2 years ago
Best ResponseYou've already chosen the best response.0ah!! yup! sorry i forgot. careless _ aaaaaccckkkkk

cuzzin
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, where does the (4x1) come from?

febylailani
 2 years ago
Best ResponseYou've already chosen the best response.0but differential is the short way. before that, there's limit. the formula is a.n.x^(n1). i.e. f(x) = 2x^3 ; a=2 ; n = 3 so, f'(x) = 2.3.x^(31) = 6x^2 cmiiw
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