Here it is, boiled down and distilled. These are the basic rules you should
know for taking derivatives.
Rule 1) The Derivative of any constant is always zero. You can know constants because they do not change with any variable. So if C never changes no matter what x (or any other variable) does, for example, then C is a constant and
= C'(x) = 0 dx
Rule 2) Multiplying a function by a constant multiplies
the resulting derivative by that same constant. Remember from rule 1 that
a constant is anything that doesn't change its value no matter what the variables do.
So if f(x) is a function whose derivative is f'(x), and
dg d(Cf) dfExamples:
= = g'(x) = C = C f'(x) dx dx dx
Rule 3) The derivative of the sum is the sum of the
This is the sum rule. In equations this means that if f(x) and g(x)
are both functions of x and
dh df dgExamples:
= h'(x) = + = f'(x) + g'(x) dx dx dx
Rule 4) If you raise x to any
power, you find the derivative by multiplying x raised to one
less than that power
by the power itself.
This is the power rule.
In equations, if you have
= f'(x) = n xn-1 dx
1 1And yes, the power rule applies even when the power is not a whole number. The power can be anything as long as it's constant.
x-1/2 = 2 2Öx
Rule 5) To find the derivative of the product of
two functions, take the first times the derivative of the second and add it
to the second times the derivative of the first.
This is the
dh dg dfExamples:
= h'(x) = f(x) + g(x) = f(x)g'(x) + g(x)f'(x) dx dx dx
_ f(x) = (x + 3)ÖxThis is the product of
1 _ f'(x) = (x + 3)
+ Öx 2Öx
f'(x) = x2 ex + 2x ex = (x2 + 2x) ex
Rule 6) To find the derivative of a quotient or ratio,
take the denominator times the derivative of the numerator, subtract from it the
numerator times the derivative of the denominator, then divide the whole thing
by the square of the denominator.
This is the
dh g(x)f'(x) - f(x)g'(x)Examples:
= h'(x) = dx g2(x)
(x + 1) f(x) =The derivative of both the numerator and denominator are both 1. So
(x - 1)
(x - 1) - (x + 1) -2 f'(x) =
= (x - 1)2 (x - 1)2
x2 + 1 g(x) =then
x3 + 1
(x3 + 1)(2x) - (x2 + 1)(3x2) g'(x) =which you can simplify using algebra if you like.
(x3 + 1)2
dtan(x) cos2(x) + sin2(x) 1
= = = sec2(x) dx cos2(x) cos2(x)
Rule 7) Use the chain rule
to find the derivative of composites.
If f(x) and g(x) are both functions of x, and
dh df dgThis means, apply the derivative of f to g(x) and multiply that by the derivative of g(x).
= h'(x) = = f'(g(x)) g'(x) dx dg dx
______ f(x) = Ö1 - x2then f is a composite of taking the square root and taking
-2x -x f'(x) =
= 2 Ög(x) Ö1 - x2
Rule 8) If you multiply the independent variable by
a constant, then the entire derivative gets multiplied by that same constant.
This is an immediate consequence of the chain rule, but it is useful
to know because it gives you a short cut. In equations, if f(x) is a
function of x and
g'(x) = k f'(kx)
Ö1 - x2
is. Applying this
rule to that derivative you find that the derivative of
Ö1 - (2x)2)
Rule 9) If you add a constant to the independent
variable, just treat the sum of the two as if it were the independent
g'(x) = f'(x + a)Examples:
f'(x) = ex+3
f'(x) = 3(x + n)2
|Derivatives of some elementary functions:|
How and when to do implicit differentiation. The chain rule applies to composites even when the function inside the composite is unknown. So, for example, if you only knew that y(x) is a function of x, and you needed to write an expression for the derivative of
f(x) = y2 - 3y + sin(y)then simply take each term and find its derivative with respect to y, then multiply that by a symbol that indicates the derivative of y. To that end, the symbol, y' does very nicely.
f'(x) = 2y y' - 3y' + cos(y) y'You could factor out a y' and get
g(x) = xy2 + y cos(x)Observe that to do this one you will have to apply the
g'(x) = [2xyy' + y2] + [-y sin(x) + y' cos(x) ]Observe that the derivative of each of the products is shown in square brackets so that you can more easily see the organization. But
g'(x) = 2xyy' + y2 - y sin(x) + y' cos(x)would be just as correct. Notice that when you take derivatives of terms that are purely functions of the independent variable, x, you don't need to multiply by x' because
The equation of x's and y's need not have an f(x) all alone on one side of the equal in order for you to use implicit differentiation. You can have complex expressions of x and y on both sides. Example:
_______ Öx2 + y2 = xy + y3To apply implicit differentiation to this equation, simply bring the appropriate rules for taking derivatives to bear onto each term as you encounter it. Remember to multiply by y' wherever it is appropriate. In this case, you apply the chain rule to the sqrt, the product rule to xy and the power rule to y3:
1 2x + 2yy'
= y + xy' + 3y2y' 2 Öx2 + y2
Going implicit with second derivatives: If you have the equation,
x3 = yy'and you wanted to apply implicit differentiation to it, you would use the power rule on the x3 and the product rule on the yy':
3x2 = yy" + (y')2Observe that as part of applying the product rule to yy' you had to take the derivative of y'. And the derivative of y' is its second deriative, y".
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