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Find the f '(x). f(x) = 2x^2 ; a=1 Help please!
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if f(x) = ax^n then f'(x) = anx^(n-1)
We are not allowed to use the short cut lol.
@Edwiener What do you mean shortcut?
The derivative has its long method and short method. We have to use f(x) -f(a)/ x-a
If so, then mention the formula: Anyway, plug the value a = 1 in, then subtract:
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Sorry, should have mentioned it.
f(x) = 2x^2 f(a) = 2a^2 ------------------------------------------------------- f(x) -f(a) 2x^2 -2a^2 2(x^2 - a^2) 2(x^2 - 1) Do you see where to go from here?
What happened to the a on the fourth step.
a = 1, plug it in!
a = 1, so I replaced 'a' with 1
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Okay, thanks!
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