{"id":2870,"date":"2024-06-10T18:41:14","date_gmt":"2024-06-10T18:41:14","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=2870"},"modified":"2024-08-05T12:40:14","modified_gmt":"2024-08-05T12:40:14","slug":"differentiation-rules-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/differentiation-rules-learn-it-2\/","title":{"raw":"Differentiation Rules: Learn It 2","rendered":"Differentiation Rules: Learn It 2"},"content":{"raw":"

The Basic Rules Cont.<\/h2>\r\n

The Sum, Difference, and Constant Multiple Rules<\/h3>\r\n

We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions. Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract, or multiply by a constant.\u00a0<\/p>\r\n

\r\n

sum, difference, and constant multiple rules<\/h3>\r\n

Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions and [latex]k[\/latex] be a constant. Then each of the following equations holds.<\/p>\r\n

 <\/p>\r\n

Sum Rule:<\/strong>\u00a0The derivative of the sum of a function [latex]f[\/latex] and a function [latex]g[\/latex] is the same as the sum of the derivative of [latex]f[\/latex] and the derivative of [latex]g[\/latex].<\/p>\r\n

[latex]\\frac{d}{dx}(f(x)+g(x))=\\frac{d}{dx}(f(x))+\\frac{d}{dx}(g(x))[\/latex];<\/div>\r\n

that is,<\/p>\r\n

for [latex]j(x)=f(x)+g(x), \\, j^{\\prime}(x)=f^{\\prime}(x)+g^{\\prime}(x)[\/latex]<\/div>\r\n
\u00a0<\/div>\r\n

Difference Rule:<\/strong>\u00a0The derivative of the difference of a function [latex]f[\/latex] and a function [latex]g[\/latex] is the same as the difference of the derivative of [latex]f[\/latex] and the derivative of [latex]g[\/latex].<\/p>\r\n

[latex]\\frac{d}{dx}(f(x)-g(x))=\\frac{d}{dx}(f(x))-\\frac{d}{dx}(g(x))[\/latex];<\/div>\r\n
\u00a0<\/div>\r\n

that is,<\/p>\r\n

for [latex]j(x)=f(x)-g(x), \\, j^{\\prime}(x)=f^{\\prime}(x)-g^{\\prime}(x)[\/latex]<\/div>\r\n
\u00a0<\/div>\r\n

Constant Multiple Rule:<\/strong>\u00a0The derivative of a constant [latex]k[\/latex] multiplied by a function [latex]f[\/latex] is the same as the constant multiplied by the derivative:<\/p>\r\n

[latex]\\frac{d}{dx}(kf(x))=k\\frac{d}{dx}(f(x))[\/latex];<\/div>\r\n

that is,<\/p>\r\n

for [latex]j(x)=kf(x), \\, j^{\\prime}(x)=kf^{\\prime}(x)[\/latex]<\/div>\r\n<\/section>\r\n
\r\n

Proof<\/strong><\/p>\r\n

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We provide only the proof of the sum rule here. The rest follow in a similar manner.<\/p>\r\n

For differentiable functions [latex]f(x)[\/latex] and [latex]g(x)[\/latex], we set [latex]j(x)=f(x)+g(x)[\/latex]. Using the limit definition of the derivative we have<\/p>\r\n

[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{j(x+h)-j(x)}{h}[\/latex]<\/div>\r\n

By substituting [latex]j(x+h)=f(x+h)+g(x+h)[\/latex] and [latex]j(x)=f(x)+g(x)[\/latex], we obtain<\/p>\r\n

[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{(f(x+h)+g(x+h))-(f(x)+g(x))}{h}[\/latex]<\/div>\r\n

Rearranging and regrouping the terms, we have<\/p>\r\n

[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}(\\frac{f(x+h)-f(x)}{h}+\\frac{g(x+h)-g(x)}{h})[\/latex]<\/div>\r\n

We now apply the sum law for limits and the definition of the derivative to obtain<\/p>\r\n

[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}(\\frac{f(x+h)-f(x)}{h})+\\underset{h\\to 0}{\\lim}(\\frac{g(x+h)-g(x)}{h})=f^{\\prime}(x)+g^{\\prime}(x)[\/latex]<\/div>\r\n

[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n

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Find the derivative of [latex]g(x)=3x^2[\/latex] and compare it to the derivative of [latex]f(x)=x^2[\/latex].<\/p>\r\n

[reveal-answer q=\"fs-id1169739286421\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169739286421\"]<\/p>\r\n

We use the power rule directly:<\/p>\r\n

[latex]g^{\\prime}(x)=\\dfrac{d}{dx}(3x^2)=3\\dfrac{d}{dx}(x^2)=3(2x)=6x[\/latex].<\/div>\r\n

Since [latex]f(x)=x^2[\/latex] has derivative [latex]f^{\\prime}(x)=2x[\/latex], we see that the derivative of [latex]g(x)[\/latex] is [latex]3[\/latex] times the derivative of [latex]f(x)[\/latex]. This relationship is illustrated in the graphs below.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"859\"]\"Two Figure 1. The derivative of [latex]g(x)[\/latex] is 3 times the derivative of [latex]f(x)[\/latex].[\/caption]\r\n\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

\r\n

Find the derivative of [latex]f(x)=2x^5+7[\/latex].<\/p>\r\n

[reveal-answer q=\"fs-id1169739064774\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169739064774\"]<\/p>\r\n

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:<\/p>\r\n

[latex]\\begin{array}{lllll}f^{\\prime}(x) & =\\frac{d}{dx}(2x^5+7) & & & \\\\ & =\\frac{d}{dx}(2x^5)+\\frac{d}{dx}(7) & & & \\text{Apply the sum rule.} \\\\ & =2\\frac{d}{dx}(x^5)+\\frac{d}{dx}(7) & & & \\text{Apply the constant multiple rule.} \\\\ & =2(5x^4)+0 & & & \\text{Apply the power rule and the constant rule.} \\\\ & =10x^4. & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

\r\n

Find the equation of the line tangent to the graph of [latex]f(x)=x^2-4x+6[\/latex] at [latex]x=1[\/latex].<\/p>\r\n

[reveal-answer q=\"fs-id1169736663036\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169736663036\"]<\/p>\r\n

To find the equation of the tangent line, we need a point and a slope. To find the point, compute<\/p>\r\n

[latex]f(1)=1^2-4(1)+6=3[\/latex].<\/div>\r\n

This gives us the point [latex](1,3)[\/latex]. Since the slope of the tangent line at [latex]1[\/latex] is [latex]f^{\\prime}(1)[\/latex], we must first find [latex]f^{\\prime}(x)[\/latex]. Using the definition of a derivative, we have<\/p>\r\n

[latex]f^{\\prime}(x)=2x-4[\/latex]<\/div>\r\n

so the slope of the tangent line is [latex]f^{\\prime}(1)=-2[\/latex]. Using the point-slope formula, we see that the equation of the tangent line is<\/p>\r\n

[latex]y-3=-2(x-1)[\/latex].<\/div>\r\n

Putting the equation of the line in slope-intercept form, we obtain<\/p>\r\n

[latex]y=-2x+5[\/latex].<\/div>\r\n

Watch the following video to see the worked solution to this example.<\/p>\r\n