H F is most polar

Do you mean y=3x or y=x3? I'll assume it's the latter. The first method to solving this is the easiest, the chain rule. Multiply the coefficient by the value of the exponent and reduce the exponent by 1. (f(x)=nAxn-1) You get y=3x2 Therefore, f(4)=3(42) f(4)=48 The longer method of solving this goes as follows: Your tangent formula is f'(x)=limh->0(f(x+h)-f(x))/h We know that f(x)=x3 so f(x+h)=(x+h)3 When we put this in the formula we have: f'(x)=limh->0((x+h)3-x3)/h f'(x)=limh->0((x+h)(x+h)(x+h)-x3)/h f'(x)=limh->0((x2+2hx+h2)(x+h)-x3)/h f'(x)=limh->0(x3+x2h+2x2h+2xh2+xh2+h3-x3)/h f'(x)=limh->0(3x2+3xh+h2) f'(x)=3x2+3x(0)+(02) f'(x)=3x2 And from there again we sub in 4 for x. f'(4)=3(42) f'(4)=48

4 Hobbits in the Fellowship

4 finger and a thumb on a hand

1. H-H 2. H-I 3. H-Br 4. H-Cl 5. H-F

Five fingers of the hand.

Do you mean h, rather than H? h is Planck's constant, f would conventionally be frequency.

4 fingers and 1 thumb on a hand

4 he's a jolly good fellow

I'm not sure if you're asking what the derivative is for f(x) = 1/[(x^2)+2] or for f(x) = [1/(x^2)]+2 so I'm gonna do the first one for now. Lemme know if you want the other one too! :)**Note: I'm not sure which terminology you're most familiar with using in class but just know that y is the same thing as/another way or writing y(x) or f(x) on the left side of the equation. Similarly, y' = y'(x) = f'(x) = d/dx(y) = dy/dx = d/dx[y(x)] = d/dx[f(x)].**So we know the First Principles from the Fundamental Theorem of Calculus defines the derivative as a limit:f'(x) = lim(hâ†’0)â¡ [f(x+h)-f(x)]/hOur function is:f(x) = 1/[(x^2)+2]Similarly:f(x+h) = 1/[((x+h)^2)+2]Simplify:f(x+h) = 1/[(x+h)(x+h)+2]f(x+h) = 1/[(x^2+2xh+h^2+2]Then we plug it into the limit and simplify:f'(x) = lim(hâ†’0)â¡ ã€–[1/(((x+h)^2)+2)] - [1/((x^2)+2))]ã€—/hf'(x) = lim(hâ†’0)â¡ ã€–[1/(((x+h)^2)+2)] - [1/((x^2)+2))]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ ã€–[1/(((x+h)^2)+2)]*[(x^2)+2]/[(x^2)+2]- [1/((x^2)+2))]*[((x+h)^2)+2]/[((x+h)^2)+2]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ ã€–[[(x^2)+2]-[((x+h)^2)+2]]/[[(x^2)+2]*[((x+h)^2)+2]]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ ã€–[[(x^2)+2]-[(x^2+2xh+h^2)+2]]/[[(x^2)+2]*[((x^2+2xh+h^2)+2]]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ ã€–[(x^2)+2-(x^2)-2xh-(h^2)-2]/[[(x^2)+2]*[((x^2+2xh+h^2)+2]]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ ã€–[(x^2)+2-(x^2)-2xh-(h^2)-2]/[(x^4)+2(x^3)h+(x^2)(h^2)+2(x^2)+2(x^2)+4xh+2(h^2)+4]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ ã€–[-2xh-(h^2)]/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ (2h)ã€–(-x-h)/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]ã€—*(1/h)f'(x) = lim(hâ†’0)â¡ (2)ã€–(-x-h)/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]ã€—Finally, plug in 0 for h:f'(x) = lim(hâ†’0)â¡ (2)ã€–(-x-(0))/[(x^4)+2(x^3)(0)+(x^2)((0)^2)+4(x^2)+4x(0)+2((0)^2)+4]ã€— f'(x) = (2)[-x/[(x^4)+4(x^2)+4]Factor the denominator:f'(x) = (2)[-x/[(x^2)+2)^2]Final answer: f'(x) = -2x/[((x^2)+2)^2]Hope this helped!! ~Casey

GREATEST common factor

fish

H f

4 hobbits in the fellowship (Horse Isle) WiseLeadMare (dun/grey)

The letter that have parallel line in it are H,I,F

hydrogen bonding. H-F...H-F...H-F...

4 Chambers of the Heart

The two American doll codes are 1 4 7 j - 4 f a e - h 7 a g , the next code is r 3 9 4 - e f t h - 4 4 a g

Half & Huff.

4 Faces on a Pyramid.

4 Horse Men of the Apocalypse technically 4 Horsemen of the Apocalypse so is 4 H of the A

4 hobbits in the fellowship (Horse Isle) by HorseyNorsey Black server

+4,+1,+2,-4,-1,-2

f = 54

lim as h->0 of (f(x+h) - f(x))/h or lim as x->a of (f(x) - f(a))/(x - a)

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