Math problem??

Question

if (x^2)[(ln^2)(x)], then what does y'=?

i'm especially confused with the ln squared...how do you multiply with that?

Answer 1

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

dC
= C'(x) = 0
dx

Examples:
The number, 14, doesn't change no matter what else varies. So the derivative of 14 is zero.
The value of π doesn't change no matter what else changes, so the derivative of π is always zero.
If a circle of fixed size has radius, r, then r is constant and the derivative of r is zero.
If a problem states that such-and-such happens at a constant rate, k, then k is a constant and the derivative of k is zero.

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 g(x) = C f(x) where C is a constant, then

dg d(Cf) df
= = g'(x) = C = C f'(x)
dx dx dx

Examples:
You know that the derivative of x2 is 2x. Then the derivative of 10x2 is 20x.
The derivative of sin(x) is cos(x). If r is constant, then the derivative of r sin(x) is r cos(x).
The derivative of ex is also ex. So the derivative of C ex, where C is constant is again C ex.

Rule 3) The derivative of the sum is the sum of the derivatives. This is the sum rule. In equations this means that if f(x) and g(x) are both functions of x and h(x) = f(x) + g(x) then

dh df dg
= h'(x) = + = f'(x) + g'(x)
dx dx dx

Examples:
If f(x) = x2 + 4x, which is a sum, then to find f'(x) you take the derivatives of the two summands, x2 and 4x, separately and take the sum of those derivatives. So f'(x) = 2x + 4.
The derivative of sin(x) + cos(x) is cos(x) - sin(x).
The derivative of ex + ln(x) is ex + 1/x.

Rule 4) If you raise x to any CONSTANT 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) = xn, where n is constant with respect to x, then

df
= f'(x) =
dx


n xn-1





Examples:

The derivative of x3 is 3x2.
The derivative of x10 is 10x9.
Since √x is the same as x1/2, the derivative of √x is

1

2


x-1/2 =


1

2√x




And 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.


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 product rule. In equations, if f(x) and g(x) are both functions of x and h(x) = f(x)g(x) then

dh dg df
= h'(x) = f(x) + g(x) = f(x)g'(x) + g(x)f'(x)
dx dx dx

Examples:
Suppose you have the function
f(x) = (x + 3)√x

This is the product of (x + 3) and √x. The derivative of (x + 3) is 1, and the derivative of √x is 1/(2√x). So
1
f'(x) = (x + 3) + √x
2√x

By the same rule, the derivative of sin(x)cos(x) is -sin2(x) + cos2(x). This is because the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x).
If y(x) = x2ex, then
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 quotient rule. In equations, if f(x) and g(x) are functions of x, and h(x) = f(x)/g(x), then

dh g(x)f'(x) - f(x)g'(x)
= h'(x) =
dx g2(x)

Examples:
The derivative of
(x + 1)
f(x) =
(x - 1)

The derivative of both the numerator and denominator are both 1. So


f'(x) =


(x - 1) - (x + 1)
=
(x - 1)2

-2

(x - 1)2




If you have
x2 + 1
g(x) =
x3 + 1

then
(x3 + 1)(2x) - (x2 + 1)(3x2)
g'(x) =
(x3 + 1)2

which you can simplify using algebra if you like.
Since tan(x) = sin(x)/cos(x), you can use the quotient rule to find the derivative of tan(x):
dtan(x) cos2(x) + sin2(x)
= =
dx cos2(x)

1
=
cos2(x)


sec2(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 h(x) = f(g(x)) (which means that you apply h to x by first applying g to x and then applying f to the result), then

dh df dg
= h'(x) = = f'(g(x)) g'(x)
dx dg dx

This means, apply the derivative of f to g(x) and multiply that by the derivative of g(x).

Examples:

If
______
f(x) = √1 - x2

then f is a composite of taking the square root and taking 1 - x2. If all you were doing were taking the square root of a variable, g, then the derivative would be 1/(2√g). But g is not just a variable here, it is a function of x. We know that the derivative of g(x) is -2x. So
-2x -x
f'(x) = =
2 √g(x) √1 - x2

By the same method, the derivative of ex2 is ex2(2x).
The derivative of
_
sin(√x)

is
_
cos(√x)

2√x

The derivative of sin2(x) is 2 sin(x) cos(x).

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) = f(kx) where k is a constant, then

g'(x) = k f'(kx)

Examples:
We know from rule 7 what the derivative of
______
√1 - x2

is. Applying this rule to that derivative you find that the derivative of
__________
√1 - (2x)2)

is
-x
2
√1 - (2x)2

If ω is a constant, then the derivative of sin(ωx) is ω cos(ωx).

Rule 9) If you add a constant to the independent variable, just treat the sum of the two as if it were the independent variable itself. So if g(x) = f(x + a), where a is constant, then

g'(x) = f'(x + a)

Examples:
We know that ex is its own derivative. So according to this rule, if f(x) = ex+3, then
f'(x) = ex+3

You know how to take the derivative of x3. So to take the derivative of f(x) = (x + n)3, where n is constant, you have
f'(x) = 3(x + n)2

These are their elementary functions:

function derivative

1
=
x

x-1


-1
=
x2

-x-2


|x|
|x| or sgn(x) where
x

sgn(x) = 1 if x > 0
sgn(x) = -1 if x < 0

xk kxk-1 (for constant k only)


√x = x1/2


1
=
2√x
1

2

x-1/2


ex ex

bx ln(b) bx (for b > 0)

1
ln(x)
x

sin(x) cos(x)

cos(x) -sin(x)

tan(x) sec2(x) = 1 + tan2(x)

cot(x) -csc2(x) = -1 - cot2(x)

sec(x) sec(x)tan(x)

csc(x) -csc(x)cot(x)

1
arcsin(x)
√1 - x2

-1
arccos(x)
√1 - x2

1
arctan(x)
1 + x2

-1
arccot(x)
1 + x2

Hope this helps

Answer 2

You'll most likely have to use the chain rule and product rule in order to take the derivative of this function. It's been about 6 years since I've done calculus, but look at the product rule and use that as a basis for your calculation, and when you take the derivative of the second term, use the chain rule where u = ln (x) and du = 1/x dx That should make the derivative findable. Hope this helps!

Answer 3

Never encountered an ln^2 before. I'd take the (ln^2 x) as (ln x)*(ln x) and use product rule.