DERIVATIVE RULESWITH EXAMPLES AND NOTES

There are many ways to denote the derivative, often depending on how the expression to be differentiated is presented. Since the derivative represents the slope of the tangent, the best notation is because it reminds us that the derivative is a slope = .

Here are the different ways of denoting the first derivative.

 given notation for derivative y y ' f (x) f ' (x) expression

Most calculus "doers" end up using y ' to denote the first derivative -- but, as mentioned above, there are other symbols which at times are more suited to the question.

In Related Rates problems, we should always use the format that signifies slope:

that is , or , to represent the change in volume per unit time.

Notation for higher derivatives

When we need to find a higher derivative (2nd, 3rd, etc.) the notation is similar to that for the first derivative -- but eventually, the "primes" become too numerous -- so we use either brackets around a number or Roman numerals to indicate the level of differentiation.

The 3rd derivative can be denoted :

 f (3)(x) f ' ' ' (x) y ' ' ' y (3)

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 Name Rule Example power f (x) = ax n so f ' (x) = nax n – 1 y = – 5x 3 , so y ' = – 15x 2 product f (x) = u v, so f '(x) = u' v + v' u y = 6x 2 \$ x 8 , so y ' = 12x ( x 8) + 6x 2 (8x 7) = 60x 9 . quotient chain y = f [g(x)], so y ' = f '[g(x)] g'(x) y = (2x 5 – 7x 2) 3, so y ' = 3(2x 5 – 7x 2) 2 (10x 4 – 14x)

notes:

power rule: applies to any term of the form ax n where n is a constant no matter if n is positive, negative, rational, or not, as long as it includes no variables.

product rule: so named since it's used on a product of 2 or more functions.
One of the functions is
u and the other is v.
In the example above: u =
6x 2 and v = x 8

quotient rule: so named since it's used on a quotient of 2 or more functions.
The numerator function is
u and the denominator function is v.

HINT: do the "v 2 " part first or you'll forget it!

In the example above: u = x 3 + 5 and v = 2x + 1

don't be fooled into thinking that is a quotient! The numerator is a constant! Not a function, therefore rewrite the function as – 7(x + 3) – 1 , then use the chain rule.

A fraction like , so it too is not a quotient.

chain rule: This is the one that causes lots of headaches.

It deals with layers of functions -- so pretend to peel the layers like peeling an onion.

f '[ g(x) ] · g '(x) means:

1st: take f ' , the outside function's derivative -- leave inside alone!
2nd: multiply by derivative of inside function.

Example:

h(x) = (–3x 2 + 5x – 1) 4 :
outside function f is ( thing ) 4 ,
inside
thing = g(x) = (3x 2 + 5x 1).

f '(x) = 4 ( thing ) 3 , and g '(x) = 6x + 5

so h '(x) = f '[ g(x) ]· g '(x)= 4 (3x 2 + 5x 1) 3 (– 6x + 5)

Notice that when we find f '(x) -- on the outside layer -- we leave g(x) alone.
We're dealing strictly with the function f (x) -- once that's done -- we find g '(x) and multiply.
So, if h (x) = ( something ) n , h '(x) = n ( something ) n – 1 (something ) '

Examples:

 function derivative – 3(x 5 + 7x 3 + 2x – 17) 15 – 45(x 5 + 7x 3 + 2x – 17) 14 (5x 4 + 21x 2 + 2) (2x + 7) 3 (4x – 3) 5 3(2x + 7) 2 (2)(4x – 3) 5 + 5(4x – 3) 4 (4)(2x + 7) 3

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Most texts teach us how to differentiate trig functions when the argument (angle) is just x.

If the argument of the trig function is u = f (x), then the derivative must include du, the derivative of the argument. And, since we can't change the argument of the trig function, we should write du at the beginning of the expression for the derivative. Otherwise, we may be tempted to multiply the du by the argument -- and we mustn't do that.

 function derivative function derivative (sin u) du cos u (cos u) – du sin u (tan u) du sec2 u (cot u) – du csc2 u (sec u) du sec u tan u (csc u) – du csc u cot u

Note1: If the function's name includes the "co" syllable (ex: cotan), the derivative is negative!

Note2: Notice the symmetry of the pairs.

For instance, (tan u)' = du sec2 u,

so (cot u)' = – du cosec2 u

It's the same for the other pair. Just add the minus sign and the "co" syllable.

since (sec u) ' = du sec u tan u

then (cosec u) ' = – du cosec u cotan u

Note: the notation for powers of the trig functions can be confusing. If y = sin 3 (7x) we're really indicating that y = (sin 7x) 3 however, this notation causes confusion since it looks like the (7x) is cubed also -- when it's not!

The derivative of y = sin 3 (7x) is y ' = (3 sin 2 (7x))( 7 cos 7x)

3 sin 2 (7x) is the derivative of (sin 7x) 3

7 cos 7x is du (7) times the derivative of sin 7x (cos 7x).

we're using the chain rule on 3 levels of functions:

level 1: thing cubed gives us 3(thing) 2

level 2: sin (something) gives us cos (something)

level 3: something (7x) gives us (something)' (7 -- in front!)

Examples

 function derivative – 5cos 4(3x+1) 20(3) cos 3 (3x+1) sin (3x+1) (chain rule) tan 2 3x sin 5x 6 tan 3x sec 2 3x sin 5x + 5cos 5x tan 2 3x (product & chain rule) 2 csc 5 x sec 2 3x 2[–5csc 4 x (csc x cot x)sec 2 3x + 6 sec 3x (sec 3x tan3x) csc 5 x

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Just as in the trig function derivatives, du represents the derivative of the argument function.

 function derivative function derivative (arcsin u) (arccos u) (arctan u) (arcsec u)

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Note1: Arctan's derivative is the only one with no root and with a plus sign

Note2: (arcsin u) ' = negative of (arccos u) '

Note3: arcsec's derivative is the wierdo in the bunch -- the order of the radicand is reversed and there's that "u" outside the radical.

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Let's prove that the derivative of y = arcsin x is

if y = arcsin x, then sin y = x -- (that is: y is the angle with a sine of x.)

now we differentiate implicitly

y ' cos y = 1 so

but, since cos 2 y + sin 2 y = 1

, since sin y = x.

Examples:

 function derivative function derivative 3 arcsin (x 2 + 5) arccos (6x – 7) arctan (3 – 2x 3 ) – 2 arcsec (e 2x + 1)

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 function derivative function derivative (e u) du e u (a u ) du a u ln a (ln u) (log a u)

see examples 2, 3, 4 and 5 in Practice exercise.

These 2 are the easiest to differentiate and yet so many students find them confusing.

Note1: exponential function: y ' = (derivative of exponent) (original function unchanged).

that's du e u

take derivative of power function (up top) and plunk it down front.

So if y = e 5 x – 3 then y ' = 5 e 5 x – 3 .

See it? The derivative of (5x 3 -- the exponent) in front of the original function.

Note 2: logarithmic function: if y = ln (thing) then

So if y = ln (5x 3 4x 2 + 3x)

Then

Note 3: Notice the difference between the derivatives of y = e u and y = a u. There's no ln a in the derivative of e u. Well, actually, there is. However, since a = e, ln a = ln e which = 1, so we don't bother to write it.

Note4: Notice the difference between the derivatives of y = ln u and y = log a u. Same as in Note3 -- the ln a is missing in the denominator when the base = e (ln u) because ln e = 1.

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Differentiate and Simplify

In this exercise, you will first see an example and then have 2 identical type questions to do.

The objective is to be as efficient as possible!! That means absolutely no extra brackets or steps. Plan ahead -- leave spaces to fill in with du's or constants as you go.

Ex. 1) y = 3 sin 2 5x so y ' = 6 (5) sin 5x cos 5x

( 6 from 2 times 3 and the 5 is the du.)

DO THESE:

 1) y = 2 cos 3 7x 2) y = 5 tan 2 3x

Ex 2) , so , ( it's du e u)

DO THESE:

 3) y = 2e 7x + 1 4) y = – 6 e – x ³ + 4

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Ex 3) y = ( 3 ) 2 x – 5, so y ' = 2(3) 2 x – 5 ln 3, (it's du a u ln a)

DO THESE:

 5) y = – (5) 7 – 3 x 6) y = 4 (½) x ² + 3x + 6

Ex 4) y = – 3 ln (x 6 – 2x 2 + 9), so

DO THESE:

 7) y = 7 ln (10x3 – x + 9) 8) y = ln (sin 2 5x)

Ex 5)

y = log 3 (4x 3 6x + 1), so

DO THESE:

 9) y = – 3 log 5 (10x3 – x + 9) 10) y = log 7 (cos 5 3x)

Ex 6) y = 2 arctan (3x 4 + 7), so

DO THESE:

 11) y = 8 arccos (3x5 + x) 12) y = – arcsec (5x4 – 9)

Ex 7) With an ugliness involving logs like this one,

rewrite y using rules of logs before differentiating.

y = ( 1/5) [ ln (x 2 6x + 3) ½ ln (5x + 7)] now differentiate

**** specify that this is y and not y ' !!!!

Reminder: Rules of Logs (logs are exponents! -- that's why they behave like this.)

log (mn) = log m + log n

log m n = n (log m)

log a 1 = 0

if log a b = y then a y = b

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 1) y = 2 cos 3 7x y ' = – 2 · 3 · 7 cos 2 7x sin 7x 2) y = 5 tan 2 3x y ' = 2 · 5 · 3 tan 3x sec 2 3x

 3) y = 2e 7x + 1 y / = 2(7)(e 7 x + 1) 4) y = – 6 e – x ³ + 4 y / = 18 x ² ( e – x ³ + 4 )

 5) y = – (5) 7 – 3 x y ' = 3 (5) 7 – 3 x ln 5 6) y = 4 (½) x ² + 3x + 6 y / = 4(2x + 3) (½) x ² + 3x ln(½)

 7) y = 7 ln (10x3 – x + 9) 8) y = ln (sin 2 5x)

 9) y = – 3 log 5 (10x3 – x + 9) 10) y = log 7 (cos 5 3x)

 11) y = 8 arccos (3x5 + x) 12) y = – arcsec (5x4 – 9)

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