Calculus Exploration 4:  Derivative Rules (D-Rules)

Constant Multiple Rule (CMR)

D*[c*f(x)] = c*D*[f(x)]

Sum of Functions Rule (SR)

D*[f(x)+g(x)] = D*[f(x)]+D*[g(x)]

Constant Function Rule (CFR)

`If for all x`, ` `*f(x) = c, ` where c is a constant number`, ` then `*D*[f(x)] = D*[c]*` = 0`

Identity Function Rule (IFR)

`If for all x`, ` `*f(x) = x, ` then `*D*[f(x)] = D*[x]*` = 1`

Power Function Rule   (PFR)

`If for all x`, ` `*f(x) = x^n, ` then `*D*[f(x)] = D*[x^n]*` = n`*x^(n-1)

Product of Functions Rule (PR)

D*[f(x)*g(x)] = D*[f(x)]*g(x)+D*[g(x)]*f(x)

Quotient of Functions Rule (QR)

D*[f(x)/g(x)] = (D*[f(x)]*g(x)-D*[g(x)]*f(x))/g(x)^2

Power Chain Rule (PowCR)

D*[f(x)^n] = n*f(x)^(n-1)*D*[f(x)]

` Note the special case: `*D*[x^n] = n*x^(n-1)*D*[x]

Exponential Rule (ExpR) & Exponential Chain Rule (ExpCR)

D*[exp(x)] = exp(x)*`   ( ExpR )`

` `*D*[exp(f(x))] = exp(f(x))*D*[f(x)]*`   ( ExpCR )`

Natural Log Rule (LnR) & Natural Log Chain Rule (LnCR)

D*[ln(x)] = 1/x, `   ( LnR )`

` `*D*[ln(f(x))] = 1/f(x)*D*[f(x)], `   ( LnCR )`

Sine Rule (SinR) & Sine Chain Rule (SinCR)

D*[sin(x)] = cos(x)*`   ( SinR )`

` `*D*[sin(f(x))] = cos(f(x))*D*[f(x)]*`   ( SinCR )`

Cosine Rule (CosR) & Cosine Chain Rule (CosCR)

D*[cos(x)] = -sin(x)*`   ( CosR )`

` `*D*[cos(f(x))] = -sin(f(x))*D*[f(x)]*`   ( CosCR )`

Tangent Rule (TanR) & Tangent Chain Rule (TanCR)

D*[tan(x)] = sec(x)^2*`   ( TanR )`

` `*D*[tan(f(x))] = sec(f(x))^2*D*[f(x)]*`   ( TanCR )`

Cotangent Rule (CotR) & Cotangent Chain Rule (CotCR)

D*[cot(x)] = -csc(x)^2*`   ( CotR )`

` `*D*[cot(f(x))] = -csc(f(x))^2*D*[f(x)]*`   ( CotCR )`

Secant Rule (SecR) & Secant Chain Rule (SecCR)

D*[sec(x)] = sec(x)*tan(x)*`   ( SecR )`

` `*D*[sec(f(x))] = sec(f(x))*tan(f(x))*D*[f(x)]*`   ( SecCR )`

Cosecant Rule (CscR) & Cosecant Chain Rule (CscCR)

D*[csc(x)] = -csc(x)*cot(x)*`   ( CscR )`

` `*D*[csc(f(x))] = -csc(f(x))*cot(f(x))*D*[f(x)]*`   ( CscCR )`


 Examples Using the Product Rule (PR)

D*[f(x)*g(x)] = D*[f(x)]*g(x)+D*[g(x)]*f(x)

Example 1.

D*[f(x)*g(x)] = D*[f(x)]*g(x)+D*[g(x)]*f(x)

f(x) = 9*x^2+7*x, ` `*g(x) = x^2-1

``

D*[f(x)*g(x)] = ``

D*[(9*x^2+7*x)*(x^2-1)] = ``

D(9*x^2+7*x)*(x^2-1)+D(x^2-1)*(9*x^2+7*x) = ``

(18*x+7)*(x^2-1)+2*x*(9*x^2+7*x) = ``

18*x^3-18*x+7*x^2-7+``

18*x^3+14*x^2 = ``

36*x^3+21*x^2-18*x-7

Example 2.

D*[f(x)*g(x)] = D*[f(x)]*g(x)+D*[g(x)]*f(x)

f(x) = x^2+3*x-5, ` `*g(x) = x^3+6*x^2-2*x+1

``

D*[f(x)*g(x)] = ``

D*[(x^2+3*x-5)*(x^3+6*x^2-2*x+1)] = ``

D(x^2+3*x-5)*(x^3+6*x^2-2*x+1)+D(x^3+6*x^2-2*x+1)*(x^2+3*x-5) = ``

(2*x+3)*(x^3+6*x^2-2*x+1)+(3*x^2+12*x-2)*(x^2+3*x-5) = ``

2*x^4+15*x^3+14*x^2-4*x+3+``

3*x^4+21*x^3+19*x^2-66*x+10 = ``

5*x^4+36*x^3+33*x^2-70*x+13


Motivation:  A Not Product Rule

`We know from the Sum Rule that `*D(x+x^2) = D(x)+D(x^2)

`So, maybe products work like this too, possibly `*D(` x`*x^2) = D(x)*D(x^2)*`??`

`But by the Power Rule`, 3*x^2*`=`*D(x^3) = D(` x`*x^2), `and `*D(x)*D(x^2) = 2*x*`.`

`So, `*D(` x`*x^2) <> D(x)*D(x^2), `and therefore the simple rule above can't be true !`

``

`And we can see the more complicated Product Rule hiding in this example:`

D(` x`*x^2)*`= 3`*x^2 = ` 1`*x^2+2*x*` x`

`             ` = D(x)*x^2+D(x^2)*x


 Examples Using the Quotient Rule (QR)

D(f(x)/g(x)) = (D(f(x))*g(x)-D(g(x))*f(x))/(g(x)^2)

Example 1.

D(f(x)/g(x)) = (D(f(x))*g(x)-D(g(x))*f(x))/(g(x)^2)

f(x) = 9*x^2+7*x, ` `*g(x) = x^2-1

``

D(f(x)/g(x)) = D((9*x^2+7*x)/(x^2-1))*`=`

(D(9*x^2+7*x)*(x^2-1)-D(x^2-1)*(9*x^2+7*x))/((x^2-1)^2) = ``

((18*x+7)*(x^2-1)-2*x*(9*x^2+7*x))/((x^2-1)^2) = ``

(18*x^3-18*x+7*x^2-7)*`+`*(-18*x^3-14*x^2)/((x^2-1)^2) = ``

(-7*x^2-18*x-7)/((x^2-1)^2)

Example 2.

D(f(x)/g(x)) = (D(f(x))*g(x)-D(g(x))*f(x))/(g(x)^2)

f(x) = x^2+3*x-5, ` `*g(x) = x^3+6*x^2-2*x+1

``

D(f(x)/g(x)) = D((x^2+3*x-5)/(x^3+6*x^2-2*x+1))*`=`

(D(x^2+3*x-5)*(x^3+6*x^2-2*x+1)-D(x^3+6*x^2-2*x+1)*(x^2+3*x-5))/((x^3+6*x^2-2*x+1)^2) = ``

((2*x+3)*(x^3+6*x^2-2*x+1)-(3*x^2+12*x-2)*(x^2+3*x-5))/((x^3+6*x^2-2*x+1)^2) = ``

(2*x^4+15*x^3+14*x^2-4*x+3)*`+`*(-3*x^4-21*x^3-19*x^2+66*x-10)/((x^3+6*x^2-2*x+1)^2) = ``

(-x^4-6*x^3-5*x^2+62*x-7)/((x^3+6*x^2-2*x+1)^2)


Motivation for QR:  Try Some Algebra and Rules You Already Know

`For g(x) nonzero, we know that`, ` g`(x)/g(x) = 1

`So, the lefthand side is just g`(x), `times `, 1/g(x), `and`

`by PR and CFR, whatever`*D(1/g(x))*`is, it must satisfy: `

D(g(x))*``(1/g(x))+D(1/g(x))*g(x) = D(1)*`= 0`

D(1/g(x))*g(x) = -D(g(x))/g(x)

D(1/g(x)) = -D(g(x))/g(x)^2

``

`Now, use this and PR again to get`*D(f(x)/g(x))*`in a similar way:`

D(f(x)*``(1/g(x))) = D(f(x))*``(1/g(x))+D(1/g(x))*f(x)

`               ` = D(f(x))/g(x)-D(g(x))/g(x)^2*f(x)

`                      ` = (D(f(x))*g(x)-D(g(x))*f(x))/g(x)^2


 Examples Using the Power Chain Rule (PowCR)

D(f(x)^n) = n*f(x)^(n-1)*D(f(x))


Example 1.

D(f(x)^n) = n*f(x)^(n-1)*D(f(x))

f(x) = 2*x^3-5*x+1, ` `*n = 10

``

D(f(x)^n) = D((2*x^3-5*x+1)^10)*`=`

10*(2*x^3-5*x+1)^9*D(2*x^3-5*x+1) = ``

10*(2*x^3-5*x+1)^9*(6*x^2-5)

Example 2.

D(f(x)^n) = n*f(x)^(n-1)*D(f(x))

f(x) = x^2+3*x-5, ` `*n = 4/3

``

D(f(x)^n) = D((x^2+3*x-5)^(4/3))*`=`

4/3*(x^2+3*x-5)^(1/3)*D(x^2+3*x-5) = ``

4/3*(x^2+3*x-5)^(1/3)*(2*x+3)

Example 3.

D(f(x)^n) = n*f(x)^(n-1)*D(f(x))

f(x) = x^6-7*x^5+8*x^3+9*x^2+1, ` `*n = 1/3

``

D(f(x)^n) = D((x^6-7*x^5+8*x^3+9*x^2+1)^(1/3))*`=`

``

1/3*(x^6-7*x^5+8*x^3+9*x^2+1)^`-`(2/3)*D(x^6-7*x^5+8*x^3+9*x^2+1) = ``

1/3*(x^6-7*x^5+8*x^3+9*x^2+1)^`-`(2/3)*(6*x^5-35*x^4+24*x^2+18*x)


Motivation for PowCR:  Try Some Algebra and Rules You Already Know

`Suppose you want to find:  `*D((x^2+3*x-5)^20)*`.  Based upon your knowledge of`

`other rules you could first multiply out, but that would be tedious, since`

(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...
(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...
(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...
(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...
(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...
(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...
(x^2+3*x-5)^20 = x^40+60*x^39+1610*x^38+25080*x^37+243295*x^36+1407672*x^35+3258690*x^34-15639660*x^33-145490505*x^32-295234920*x^31+1353572844*x^30+7851690720*x^29+746643570*x^28-86198131440*x^27-1367...

``

`Instead, you might try to use PR by first considering a simpler case: `*D((x^2+3*x-5)^2)

`Then, by PR,  `*D((x^2+3*x-5)^2) = D((x^2+3*x-5)*` `(x^2+3*x-5))

`             ` = D(x^2+3*x-5)*(x^2+3*x-5)*` `+` D`(x^2+3*x-5)*(x^2+3*x-5)

`` = `2`(x^2+3*x-5)*D(x^2+3*x-5)

`` = `2`(x^2+3*x-5)*(2*x+3)

``

`Let's see if this pattern persists when we try n = 3.  By PR and the above, we have`

D((x^2+3*x-5)^3) = D((x^2+3*x-5)*` `(x^2+3*x-5)^2)

`             ` = D(x^2+3*x-5)*(x^2+3*x-5)^2*` `+` D`((x^2+3*x-5)^2)*(x^2+3*x-5)

`             ` = (2*x+3)*(x^2+3*x-5)^2*` `+` 2`(x^2+3*x-5)*(2*x+3)*(x^2+3*x-5)

`             ` = (2*x+3)*(x^2+3*x-5)^2*` `+` 2`(2*x+3)*(x^2+3*x-5)^2

`` = `3 `(x^2+3*x-5)^2*(2*x+3)

``

`Then so far the pattern is: `

D((x^2+3*x-5)^2) = `2 `(x^2+3*x-5)^` 1`*D(x^2+3*x-5)

D((x^2+3*x-5)^3) = `3 `(x^2+3*x-5)^2*D(x^2+3*x-5)

``

`Apparently, we can extend this pattern to answer our original question,`

`which is much less tedious!`

D((x^2+3*x-5)^20) = `20 `(x^2+3*x-5)^19*D(x^2+3*x-5)

`                        ` = `20 `(x^2+3*x-5)^19*(2*x+3)

``

Dr. John Pais, Mathematics Department-MICDS

E-mail: pais@micds.org or pais@kinetigram.com

URL:  http://kinetigram.com/micds