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We now apply the sum law for limits and the definition of the derivative to obtain

j ( x ) = lim h 0 ( f ( x + h ) f ( x ) h ) + lim h 0 ( g ( x + h ) g ( x ) h ) = f ( x ) + g ( x ) .

Applying the constant multiple rule

Find the derivative of g ( x ) = 3 x 2 and compare it to the derivative of f ( x ) = x 2 .

We use the power rule directly:

g ( x ) = d d x ( 3 x 2 ) = 3 d d x ( x 2 ) = 3 ( 2 x ) = 6 x .

Since f ( x ) = x 2 has derivative f ( x ) = 2 x , we see that the derivative of g ( x ) is 3 times the derivative of f ( x ) . This relationship is illustrated in [link] .

Two graphs are shown. The first graph shows g(x) = 3x2 and f(x) = x squared. The second graph shows g’(x) = 6x and f’(x) = 2x. In the first graph, g(x) increases three times more quickly than f(x). In the second graph, g’(x) increases three times more quickly than f’(x).
The derivative of g ( x ) is 3 times the derivative of f ( x ) .
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Applying basic derivative rules

Find the derivative of f ( x ) = 2 x 5 + 7 .

We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:

f ( x ) = d d x ( 2 x 5 + 7 ) = d d x ( 2 x 5 ) + d d x ( 7 ) Apply the sum rule. = 2 d d x ( x 5 ) + d d x ( 7 ) Apply the constant multiple rule. = 2 ( 5 x 4 ) + 0 Apply the power rule and the constant rule. = 10 x 4 . Simplify.
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Find the derivative of f ( x ) = 2 x 3 6 x 2 + 3 .

f ( x ) = 6 x 2 12 x .

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Finding the equation of a tangent line

Find the equation of the line tangent to the graph of f ( x ) = x 2 4 x + 6 at x = 1 .

To find the equation of the tangent line, we need a point and a slope. To find the point, compute

f ( 1 ) = 1 2 4 ( 1 ) + 6 = 3 .

This gives us the point ( 1 , 3 ) . Since the slope of the tangent line at 1 is f ( 1 ) , we must first find f ( x ) . Using the definition of a derivative, we have

f ( x ) = 2 x 4

so the slope of the tangent line is f ( 1 ) = −2 . Using the point-slope formula, we see that the equation of the tangent line is

y 3 = −2 ( x 1 ) .

Putting the equation of the line in slope-intercept form, we obtain

y = −2 x + 5 .
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Find the equation of the line tangent to the graph of f ( x ) = 3 x 2 11 at x = 2 . Use the point-slope form.

y = 12 x 23

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The product rule

Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule    does not follow this pattern. To see why we cannot use this pattern, consider the function f ( x ) = x 2 , whose derivative is f ( x ) = 2 x and not d d x ( x ) · d d x ( x ) = 1 · 1 = 1 .

Product rule

Let f ( x ) and g ( x ) be differentiable functions. Then

d d x ( f ( x ) g ( x ) ) = d d x ( f ( x ) ) · g ( x ) + d d x ( g ( x ) ) · f ( x ) .

That is,

if j ( x ) = f ( x ) g ( x ) , then j ( x ) = f ( x ) g ( x ) + g ( x ) f ( x ) .

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Proof

We begin by assuming that f ( x ) and g ( x ) are differentiable functions. At a key point in this proof we need to use the fact that, since g ( x ) is differentiable, it is also continuous. In particular, we use the fact that since g ( x ) is continuous, lim h 0 g ( x + h ) = g ( x ) .

By applying the limit definition of the derivative to j ( x ) = f ( x ) g ( x ) , we obtain

Practice Key Terms 7

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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