Calculus help

M+M

M+M

Active member
taking first and second derivatives of things like (3/x^2)....how to do that and then the opposite with things like (x^4/4) help is greatly appretiated thanks
 
general power rule for 1st derivatives

X^n = nX^n-1 so if you have X^4 the derivative is 4x^3, for second deriv you just repeat ,so 4x^3 would go to 12x^2, and so on until it becomes a constant which goes to zero

Antiderivatives: general formula: x^n goes to 1/n+1x^n+1 + C, so the anti deriv of X^4 would then be (1/5)x^5+C where C is a constant.

hope this helps
 
it's been a little while since my last calc. class but i think it goes something like this:

1/x^n is the same thing as x^-n

so if you have (3/(x^2)) you can write it like 3x^-2

then you take the first derivative like you normally would:

3(-2)x^(-2-1)

and you end up with -6x^-3 or (-6/(x^3))

this is different from ((x^4)/4). Here, x is in the numerator so you can just derive it right away:

(4(x^3))/4 or x^3.

i'm pretty sure that's how it goes.

 
when taking derivatives, multiply the base times the power, then write that with the exponent minus one. for example, 4x^3 turns into 12x^2. the derivative of x is simply 1, and the derivatives of constants, like 3 or 12, is zero.
 
yeah, theyve got you pretty much covered there. when taking antiderivatives dont forget to add the C at the end to represent a constant
 
i got an A- in honors calc first quarter...but its my senior year, and i kinda stoped going to classes and wat not already, probably shouldnt have done that
 
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