The Derivative rules are used to find the derivative of a term or terms without having to simplify first.
Here is an example and proof that the derivative rules work.
Using derivative rulesSimplifying first
Y=(2x5+5)(5x2+2)Y=(2x5+5)(5x2+2)
y'=(10x4)(5x2+2)+(2x5+5)(10x)y=10x7+4x5+20x6+50x
y'=50x6+20x4+20x6+50xy'=70x6+20x4+50x
y'=70x6+20x4+50x
Both methods give you the same answer.
The exponent rule is use to find the derivative of a term with an exponent.
When using the exponential rule you take the exponent and multiply it by it’s base and subtract one from the exponent.
When simplifying there are a few rules to remember.
- No Negative exponents
- No fractional exponents
- No radicals in denominator
The derivative of x³ would be 3x² and the second derivative would be 6x.
Example:
Y=x³+5x²+10 The derivative of a constant is 0 so any real numbers become 0
y'=3x²+10x x° is 1
y''=6x+10
y'''=6
y''''=0
y=4x³-2x²+2x-27
y'=12x²-4x+2
y''=24x-4
y'''=24
y''''=0
The Product rule is used when you need to find the derivative of two terms that are going to be multiplied together.
The product rule is you take the derivative of the first term then multiply it by the second term then add the second term multiplied by the first term.
(Derivative of the 1st Term)(2nd Term)+(Derivative of the 2nd Term)(1st Term)
Example
(x²+3x)(x+4)[(2x+1)(3x+1)](4x+1)
Use Product rule[2(3x+1)+3(2x+1)](4x+1)+4[(2x+1)(3x+1)]
(2x-3)(x+4)+(1)(x²-3x)[6x+2+6x+3]+4(6x²+3x+1)
Simplify(12x+5) (4x+1)
(2x²+5x-12)+(x²-3x)48x²+32x-15+24x²+20x+4
y'=3x²+2x-1272x²+52x+9
The quotient rule is used to find the derivative of one term divided by another with a denominator that is at least a binomial.
(derivative of numerator)(denominator)-(derivative of denominator)(numerator)
(Denominator)²
y=x²+x+1
x²-x-1
y'=(2x+1)(x²-x-1)-(2x-1)(x²+x+1)
(x²-x-1)²
y'=(2x³-2x²-2x)(x²-x-1)-(2x³+2x²+2x)(-x²-x-1)
(x²-x-1)²
y'=2x³-x²-3x-1-2x³+x²+x-1
(x²-x-1)²
The chain rule is used when you need to find the derivative of a binomial term with an exponent.
Derivative with respect to the bracket times the derivative of what is in the bracket.
y=(-x²+4x-2)²y=(3x5-x3/2)4
y'=2(-x²+4x-2)(-2x+4)y'=4(3x5-x3/2)3(15x4-3/2x1/2)
y'=(-4x+8)(-x²+4x-2)y'=(60x4-6(x))(3x5-x3/2)3
1)Y=x4-x2
2)Y=-2x-4
3)Y=-2x+3
4)Y=x3+2x2+x
5)Z=(W+2)(W+3)
W+5
6)W= w+2
w2+8w+15
7)U=(3v-1)2-1
(3v-1)2+1
8)m= ___1____
n2+3n-1
9)y=(2x+1)6
10)y=(5x+6)-3
11)y=(4-9x2)30
12)y=(4x-3x2)10
13)y=(2x+x4)(2x+x4)
14)y=(2x-1)(4x2-6x-3)
15)p(x)=f(x)d
1)2)3)4)
Y=x4-x2y=-2x-4y=-2x+3y=x3+2x2+x
y'=4x3-2xy'=8x-5y'=-2y'=3x2+4x+1
y''=12x2-2y'=8+1y'=0y'=6x+4
y'''=24x 1 x5y'=6
y''''=24y'=8y'=0
y'''''=0 x5
5)Z=(W+2)(W+3) Z'=w2+3w+2w+6 = 2w+5(w+5)-(w2+3w+2w+6) =
W+5w+5w+5
2w2+10w+5w+25-w2-3w-2w-6 = w2+10w+19
(w+5)2 (w+5)2
6) W= w+2= w2+8w+15-2w+8(w+2) = w2+8w+15-2w2-4w-8w-16 =
w2+8w+15(w2+8+15)2(w2+8w+15)
w'= -w2-4w-1
(w2+8w+18)2
1)U=(3v-1)2-1 U'= 9v2-6v = 18v-6(9v2-6v+2)-18v-6(9v2-6v) = 36v-12 _
(3v-1)2+1 9v2-6v+2 (9v2-6v+2)2 [(3v-1)2+1]2
2)m= ___1____ m'= 0(n2+3n-1)-2n+3(1) = __-2n+3__
n2+3n-1 (n2+3n-1)2 (n2+3n-1)2
9)10)11)12)
y=(2x+1)6 y=(5x+6)-3y=(4-9x2)30Y=(4x-3x2)10
y'=6(2x+1)5(2) y'=-3(5x+6)-4(5)y'=30(4-9x2)29(18x)y'=10(4x-3x)9(4-6x)
y=12(2x+1)5=-15(5x+6)=540x(4-9x2)29=(40-60x)(4x-3x)9
13)14)
y=(2x+x4)(2x+x4)y=(2x-1)(4x2-6x-3)
y'=(2+4x3)(2x+x4)+(2+4x3)(2x+x4)y'=2(4x2-6x-3)+(8x-6)(2x-1)
= 2(2+4x3)(2x+x4)=8x2-12-6+16x2-20x+6
=24x2-32x
15)
p(x)=f(x)d
p'(x)=f '(x)d+d'(f(x))
1) y=x3+4x2-27
2)y=(2x2+7)(6x3-2x)
3)y=(14x-5+7)3
4)p=2q+1
2q-1
5)y=___t2__
(t3+1)2
1)2)
y=x3+4x2-27y=(2x2+7)(6x3-2x)
y'=3x2+8xY'=4x(6x3-2x)+(18x2-2)(2x2+7)
y''=6x =24x4-8x2+36x4+126x2-4x2-14
=60x4+114x2-14
3) y=(14x-5+7)3 Y'= 3(14x-5+7)2(-70x-6) = (14x-5+7)2(-210x-6)
4) p=2q+1 P'=(2)(2q+1)-(2)(2q-1)=(2+4q)-(4q-2)=__4___
2q-1 (2q+1)2 (2q+1)2 (2q+1)2
5)
y=__ t2___= y'=2t(t3+1)2-2(t3+1)(3t2)(t2) = 2t(t3+1)2-6t4(t3+1) = 2t(t3+1)-6t4 = 2t4+2t-6t
(t3+1)2 (t3+1)4 (t3+1)4 (t3+1)3 (t3+1)3
= 4t4+2t-6t4 = -4t4+2t
(t3+1)3 (t3+1)3
- The-derivative-of-