derivatives?

i just had my derivative exam last monday…

deriv(x^2)

= lim (x+Δx)^2 – f(x^2)/ Δx

x->0

= lim [x^2 + 2xΔx + (Δx)^2] – x^2/Δx

x->0

cancel x^2

what is left is:

lim Δx(2x + Δx)/Δx

cancel Δx

= lim 2x + Δx

x-> 0

which means

the dy/dx of x^2 is

2x…

2. f(x)=x^3 evaluate [f(x+Δx) – f(x)]/Δx

deriv (x^3)

= lim (x+Δx)^3 – f(x^3)/ Δx

x->0

lim (x^3 +3x^2Δx + 3x(Δx)^2 + (Δx)^3 – x^3/Δx

x->0

cancel x^3

=lim 3x^2Δx + 3x(Δx)^2 + (Δx)^3/Δx

x ->0

=factor it out

= lim Δx[3x^2 + 3xΔx + (Δx)^2/Δx

x->0

cancel Δx

left is:

= 3x^2

since x approaches to zero…

okay…hope it helps…

try to do it… 🙂

you can try differential and integral calculus by feliciano and uy… 🙂

-kimi-

f(2+Δx) means replace each x with “2+Δx”

f(2+Δx) = (2+Δx)^2 – (2+Δx) + 1

= 4 + 4Δx + (Δx)^2 – 2 – Δx + 1

= (Δx)^2 + 3Δx + 3

f(2) = 2^2 – 2 + 1

f(2) = 4-2+1

f(2) = 3

Now subtract

f(2+Δx) – f(2)

= (Δx)^2 + 3Δx + 3 – 3

= (Δx)^2 + 3Δx

Now divide by Δx

((Δx)^2 + 3Δx) / Δx

= (Δx)^2 / Δx + 3Δx/Δx

= Δx + 3

2) f (x + Δx) means replace every x with “x + Δx”

f (x + Δx) = (x + Δx)^3

= (x + Δx)(x + Δx)(x + Δx)

= (x^2 + 2xΔx + (Δx)^2)(x + Δx)

= x^3 + 2x^2 Δx + x(Δx)^2 + x^2 Δx + 2x(Δx)^2 + (Δx)^3

= x^3 + 3x^2 Δx + 3x (Δx)^2 + (Δx)^3

Now subtract f(x)

f (x + Δx) – f(x)

= x^3 + 3x^2 Δx + 3x (Δx)^2 + (Δx)^3 – x^3

= 3x^2 Δx + 3x (Δx)^2 + (Δx)^3

Now divide by Δx

(3x^2 Δx + 3x (Δx)^2 + (Δx)^3) / Δx

= 3x^2 + 3xΔx + (Δx)^2