# Product Rule Problems Worked Out | Calculus Derivative Examples

The product rule in differential calculus is one of the first techniques students learn for taking derivatives. Once you get the hang of using the product rule, taking derivatives of complicated functions is a snap. The rule can seem tricky at first, but with the examples below you can see how the rule is applied to many different kinds of problems.

What is the product rule? If a function h(x) is the product of two functions f(x) and g(x), in other words, h(x) = f(x)g(x), then the derivative of h(x) is given by the formula

h* '*(x) = f

*(x)g(x) + f(x)g*

**'***(x).*

**'**In other words, you take the derivative of the first factor function and multiply it by the second factor function, then take the derivative of the second factor function and multiply it by the first factor function, and then add them together. The rule can even be extended for functions that are products of more than two functions. For instance, if K(x) = a(x)b(x)c(x), the K* '*(x) is given by

K* '*(x) = a

*(x)b(x)c(x) + a(x)b*

**'***(x)c(x) + a(x)b(x)c*

**'***(x).*

**'**## Max of h(x) = (x^n)e^(-x) for All n > 0

To find the maximum value of a single-variable function, you set the derivative equal to zero, solve for x, and then test whether the values of x are local or global maxes or mins. In this example, if h(x) = (x^n)e^(-x), the two factor functions are f(x) = x^n and g(x) = e^(-x). Here, the unknown n is treated as a constant parameter. Thus we have

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

= [nx^(n-1)]e^(-x) + (x^n)(-1)e^(-x)

= [e^(-x)][nx^(n-1) - x^n]

Setting the derivative equal to zero yields the equation

0 = [e^(-x)][nx^(n-1) - x^n]

Since the function e^(-x) is never zero, this term can be factored out from both sides of the equation, leaving us with a simpler equation.

0 = nx^(n-1) - x^n

nx^(n-1) = x^n

x = 0 or x = n

Of the two solutions x = 0 is a minimum. Why? Because the function h(x) is non-negative for all values of x, and h(0) = 0, the minimum value. This means that x = n is a local max. And since h(n) = (n/e)^n, the coordinates of the max are (n, (n/e)^n). Here is a graph of the function for n = 3 and n = 4.

## Find the Max of h(x) = Ln(x)/x

This might look like a job for the quotient rule since h(x) is of the form f(x)/g(x). However, all quotient functions can be written equivalently as products. If the factor functions are f(x) = Ln(x) and g(x) = 1/x, then h(x) = f(x)g(x). Now the derivative is

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

= (1/x)*(1/x) + Ln(x)*(-1/x^2)

= 1/x^2 - Ln(x)/x^2

= [1 - Ln(x)]/x^2

Setting it equal to zero gives us

0 = [1 - Ln(x)]/x^2

The function h(x) is only defined for positive values of x, and for all positive values of x the expression x^2 is positive. Therefore, if we multiply both sides of the equation by x^2 we get

0 = 1 - Ln(x)

1 = Ln(x)

e = x

Since this function has only one critical point and the question asked for the max, the maximum value must be at x = e. Since h(e) = 1/e, the coordinates of the max are (e, 1/e). This can be seen from the graph of the function.

## Find the Equation of the Tangent Line on h(x) = x*sqrt(x + 3) at the Point (6, 18)

To find the equation of the tangent line at a given point on a curve, we first need to find the slope of the line. The slope is given by the value of the derivative at that point. Since h(x) = x*sqrt(x+3) = x*(x+3)^(1/2), we have

h'(x) = (x+3)^(1/2) + x*[(1/2)(x+3)^(-1/2)] = (1.5x + 3)/sqrt(x+3)

To find the value of the derivative at the point (6, 18), we plug x = 6 into the expression above. This gives us

h'(6) = (1.5*6 + 3)/sqrt(6 + 3) = 4

So the slope of the tangent line is 4. We also know that the point (6, 18) lies on the line, by definition of tangency. The rest of the work is just middle school algebra. The point slope form (y - 18) = 4(x - 6) simplifies to **y = 4x - 6**. This is the equation of the tangent line.

## On What Interval Is h(x) = cos(x) + x*sin(x) Increasing?

A function is increasing when its first derivative is positive. If h(x) = cos(x) + x*sin(x), then the first derivative is

h'(x) = -sin(x) + sin(x) + x*cos(x)

= x*cos(x)

To solve the inequality equation

0 < x*cos(x)

we need to understand where the functions a(x) = x and b(x) = cos(x) are greater than 0 and less than 0. For a(x) = x, the function is positive whenever x is positive, and negative whenever x is negative. For the periodic function b(x) = cos(x), the function is positive on the open intervals ... (-5π/2, -3π/2), (-π/2, π/2), (3π/2, 5π/2), (7π/2, 9π/2),... and negative on the open intervals in between.

Since positive times positive equals positive and negative times negative equals positive, this means that the derivative h'(x) is positive on the intervals

... (-7π/2, -5π/2), (-3π/2, -π/2), (0, π/2), (3π/2, 5π/2),...

## Comments

trying to work out derivative of sin(x)^2 by writing it as sin(x)*sin(x) and using product rule. I get 2*sin(x)*cos(x) but the book says the answer is sin(2x). How does the 2 get inside?