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Ms. Buhler @Trinity
Product Rule
f’g + fg’
Quotient Rule
(gf’ - fg’) / g2
Chain Rule (ex: d/dx(2x-1)3)
(inside)’(outside(inside))’
Example: (2x-1)3 → 2 × 3(2x-1)2
Antiderivative (ex: ∫(4x3))
∫(4x3) → d/dx 4ax4 = 4a(4x3)
16ax3 = 4x3; a = ¼
4(1/4)x4 = x4
Proving Product Rule
(fg)’
limx→a (f(x)g(x) - f(a)g(a)) / (x-a)
limx→a(f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)) / (x-a)
limx→a(f(x)(g(x) - g(a)) + g(a)(f(x) - f(a)) / (x-a)
f(a) * limx→a (g(x) - g(a)) / (x-a) + g(a) * limx→a (f(x)-f(a)) / (x-a)
f(a) * g’(a) + g(a) * f’(a)
Proving Quotient Rule
(f/g)’
limx→a (f(x)/g(x) - f(a)/g(a)) / (x-a)
limx→a (f(x)g(a) - f(a)g(x)) / g(x)g(a)) / (x-a)
limx→a 1/g(x)g(a) * limx→a (f(x)g(a) - f(a)g(x)) / (x-a)
limx→a 1/g(x)g(a) * limx→a (f(x)g(a) - f(a)g(a) + f(a)g(a) - f(a)g(x)) / (x-a)
limx→a 1/g(x)g(a) * limx→a (g(a)(f(x) - f(a)) + f(a)(g(a) - g(x))) / (x-a)
limx→a 1/g(x)g(a) * g(a) * limx→a (f(x) - f(a)) / (x-a) + f(a) * limx→a (g(a) - g(x)) / (x-a)
1/g(a)2 * g(a) * f’(a) - f(a) * g’(a)
(g(a) * f’(a) - f(a) * g’(a)) / (g(a))2
Antiderivative of acceleration
Antiderivative of velocity
Derivative of velocity
Velocity
Position
Acceleration
How to find local extrema?
Find first derivative of function and set equal to 0 to get the locations of the local extrema (x-value). Use two values above and below the resultant value(s) to find out which one they identify as. Positive f’ to negative f’ means a local maximum; the opposite means a local minimum. If you need to find a specific point, plug the x-value back into the original equation to get its appropriate y-value.
How to find inflection points?
Find second derivative of function and set equal to 0 to get the locations of the inflection points (x-value). Use two values above and below the resultant value(s) to find out whether it changes from positive to negative — if concavity changes, it is an inflection point!
Positive f’’ means a positive concavity, like a smile. Negative f’’ means a negative concavity, like a frown.