{"id":3332,"date":"2024-06-20T14:43:25","date_gmt":"2024-06-20T14:43:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=3332"},"modified":"2024-08-05T12:47:44","modified_gmt":"2024-08-05T12:47:44","slug":"the-chain-rule-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-chain-rule-learn-it-3\/","title":{"raw":"The Chain Rule: Learn It 3","rendered":"The Chain Rule: Learn It 3"},"content":{"raw":"

Applying the Chain Rule Multiple Times<\/h2>\r\n

We can integrate the chain rule with other differentiation rules to handle functions composed of multiple functions. When differentiating compositions that involve three or more functions, it\u2019s often necessary to apply the chain rule multiple times. This allows us to systematically derive the derivative without memorizing complex formulas, as the application of the chain rule can be repeated as needed.<\/p>\r\n

In a general case, consider a function [latex]k(x)=h(f(g(x)))[\/latex].<\/p>\r\n

First, apply the chain rule to find:<\/p>\r\n

[latex]k^{\\prime}(x)=\\frac{d}{dx}(h(f(g(x))))=h^{\\prime}(f(g(x))) \\cdot \\frac{d}{dx}(f(g(x)))[\/latex]<\/div>\r\n

Applying the chain rule once again gives us:<\/p>\r\n

[latex]k^{\\prime}(x)=h^{\\prime}(f(g(x)))f^{\\prime}(g(x))g^{\\prime}(x)[\/latex]<\/div>\r\n
\r\n

chain tule for a composition of three functions<\/h3>\r\n

For all values of [latex]x[\/latex] for which the function is differentiable, if<\/p>\r\n

[latex]k(x)=h(f(g(x)))[\/latex],<\/div>\r\n

then<\/p>\r\n

[latex]k^{\\prime}(x)=h^{\\prime}(f(g(x)))f^{\\prime}(g(x))g^{\\prime}(x)[\/latex]<\/div>\r\n

In other words, we are applying the chain rule twice.<\/p>\r\n<\/section>\r\n

\r\n

Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we always work from the outside in, taking one derivative at a time.<\/em><\/p>\r\n<\/section>\r\n

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Find the derivative of [latex]k(x)=\\cos^4 (7x^2+1)[\/latex]<\/p>\r\n

[reveal-answer q=\"fs-id1169736658640\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169736658640\"]<\/p>\r\n

First, rewrite [latex]k(x)[\/latex] as<\/p>\r\n

[latex]k(x)=(\\cos(7x^2+1))^4[\/latex]<\/div>\r\n

Then apply the chain rule several times.<\/p>\r\n

[latex]\\begin{array}{lllll}k^{\\prime}(x) & =4(\\cos(7x^2+1))^3(\\frac{d}{dx}\\cos(7x^2+1)) & & & \\text{Apply the chain rule.} \\\\ & =4(\\cos(7x^2+1))^3(\u2212\\sin(7x^2+1))(\\frac{d}{dx}(7x^2+1)) & & & \\text{Apply the chain rule.} \\\\ & =4(\\cos(7x^2+1))^3(\u2212\\sin(7x^2+1))(14x) & & & \\text{Apply the chain rule.} \\\\ & =-56x \\sin(7x^2+1) \\cos^3 (7x^2+1) & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n\r\n[\/hidden-answer]<\/section>\r\n
\r\n

Don't forget that [latex]{\\cos }^{n}(t)[\/latex] is a commonly used shorthand notation for [latex]{\\left(\\cos \\left(t\\right)\\right)}^{n}[\/latex]. When we write it without the shorthand notation, we can clearly see why the chain rule is necessary in these situations.<\/p>\r\n<\/section>\r\n

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[ohm_question hide_question_numbers=1]206008[\/ohm_question]<\/p>\r\n<\/section>\r\n

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A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)= \\sin (2t)+ \\cos (3t)[\/latex]. What is the velocity of the particle at time [latex]t=\\frac{\\pi}{6}[\/latex]?<\/p>\r\n

[reveal-answer q=\"fs-id1169736593621\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169736593621\"]<\/p>\r\n

To find [latex]v(t)[\/latex], the velocity of the particle at time [latex]t[\/latex], we must differentiate [latex]s(t)[\/latex]. Thus,<\/p>\r\n

[latex]v(t)=s^{\\prime}(t)=2 \\cos(2t)-3 \\sin(3t)[\/latex].<\/div>\r\n

 <\/p>\r\n

Substituting [latex]t=\\frac{\\pi}{6}[\/latex] into [latex]v(t)[\/latex], we obtain [latex]v(\\frac{\\pi}{6})=-2[\/latex].<\/p>\r\n

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