{"id":254,"date":"2023-09-20T22:48:36","date_gmt":"2023-09-20T22:48:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-chain-rule-using-leibnizs-notation\/"},"modified":"2024-08-05T01:58:55","modified_gmt":"2024-08-05T01:58:55","slug":"the-chain-rule-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-chain-rule-learn-it-4\/","title":{"raw":"The Chain Rule: Learn It 4","rendered":"The Chain Rule: Learn It 4"},"content":{"raw":"

The Chain Rule Using Leibniz\u2019s Notation<\/h2>\r\n

As with other derivatives that we have seen, we can express the chain rule using Leibniz\u2019s notation. This notation for the chain rule is used heavily in physics applications.<\/p>\r\n

For [latex]h(x)=f(g(x))[\/latex], let [latex]u=g(x)[\/latex] and [latex]y=h(x)=g(u)[\/latex]. Thus,<\/p>\r\n

[latex]h^{\\prime}(x)=\\frac{dy}{dx}, \\, f^{\\prime}(g(x))=f^{\\prime}(u)=\\frac{dy}{du}[\/latex], and [latex]g^{\\prime}(x)=\\frac{du}{dx}[\/latex]<\/div>\r\n

Consequently,<\/p>\r\n

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

chain rule using Leibniz\u2019s notation<\/h3>\r\n

If [latex]y[\/latex] is a function of [latex]u[\/latex], and [latex]u[\/latex] is a function of [latex]x[\/latex], then<\/p>\r\n

[latex]\\dfrac{dy}{dx}=\\dfrac{dy}{du} \\cdot \\dfrac{du}{dx}[\/latex]<\/div>\r\n<\/section>\r\n
\r\n

Find the derivative of [latex]y=\\left(\\dfrac{x}{3x+2}\\right)^5[\/latex]<\/p>\r\n

[reveal-answer q=\"fs-id1169739264248\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169739264248\"]<\/p>\r\n

First, let [latex]u=\\frac{x}{3x+2}[\/latex]. Thus, [latex]y=u^5[\/latex]. Next, find [latex]\\frac{du}{dx}[\/latex] and [latex]\\frac{dy}{du}[\/latex].<\/p>\r\n

Using the quotient rule,<\/p>\r\n

[latex]\\frac{du}{dx}=\\frac{2}{(3x+2)^2}[\/latex]<\/div>\r\n

and<\/p>\r\n

[latex]\\frac{dy}{du}=5u^4[\/latex]<\/div>\r\n

Finally, we put it all together.<\/p>\r\n

[latex]\\begin{array}{lllll}\\frac{dy}{dx} & =\\frac{dy}{du} \\cdot \\frac{du}{dx} & & & \\text{Apply the chain rule.} \\\\ & =5u^4 \\cdot \\frac{2}{(3x+2)^2} & & & \\text{Substitute} \\, \\frac{dy}{du}=5u^4 \\, \\text{and} \\, \\frac{du}{dx}=\\frac{2}{(3x+2)^2}. \\\\ & =5(\\frac{x}{3x+2})^4 \\cdot \\frac{2}{(3x+2)^2} & & & \\text{Substitute} \\, u=\\frac{x}{3x+2}. \\\\ & =\\frac{10x^4}{(3x+2)^6} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.<\/p>\r\n

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

\r\n

Find the derivative of [latex]y= \\tan (4x^2-3x+1)[\/latex]<\/p>\r\n

[reveal-answer q=\"fs-id1169736661260\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169736661260\"]<\/p>\r\n

First, let [latex]u=4x^2-3x+1[\/latex]. Then [latex]y= \\tan u[\/latex]. Next, find [latex]\\frac{du}{dx}[\/latex] and [latex]\\frac{dy}{du}[\/latex]:<\/p>\r\n

[latex]\\frac{du}{dx}=8x-3[\/latex] and [latex]\\frac{dy}{du}=\\sec^2 u[\/latex].<\/div>\r\n

Finally, we put it all together.<\/p>\r\n

[latex]\\begin{array}{lllll}\\frac{dy}{dx} & =\\frac{dy}{du} \\cdot \\frac{du}{dx} & & & \\text{Apply the chain rule.} \\\\ & = \\sec^2 u \\cdot (8x-3) & & & \\text{Use} \\, \\frac{du}{dx}=8x-3 \\, \\text{and} \\, \\frac{dy}{du}= \\sec^2 u. \\\\ & = \\sec^2 (4x^2-3x+1) \\cdot (8x-3) & & & \\text{Substitute} \\, u=4x^2-3x+1. \\end{array}[\/latex]<\/div>\r\n
[\/hidden-answer]<\/div>\r\n<\/section>","rendered":"

The Chain Rule Using Leibniz\u2019s Notation<\/h2>\n

As with other derivatives that we have seen, we can express the chain rule using Leibniz\u2019s notation. This notation for the chain rule is used heavily in physics applications.<\/p>\n

For [latex]h(x)=f(g(x))[\/latex], let [latex]u=g(x)[\/latex] and [latex]y=h(x)=g(u)[\/latex]. Thus,<\/p>\n

[latex]h^{\\prime}(x)=\\frac{dy}{dx}, \\, f^{\\prime}(g(x))=f^{\\prime}(u)=\\frac{dy}{du}[\/latex], and [latex]g^{\\prime}(x)=\\frac{du}{dx}[\/latex]<\/div>\n

Consequently,<\/p>\n

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

chain rule using Leibniz\u2019s notation<\/h3>\n

If [latex]y[\/latex] is a function of [latex]u[\/latex], and [latex]u[\/latex] is a function of [latex]x[\/latex], then<\/p>\n

[latex]\\dfrac{dy}{dx}=\\dfrac{dy}{du} \\cdot \\dfrac{du}{dx}[\/latex]<\/div>\n<\/section>\n
\n

Find the derivative of [latex]y=\\left(\\dfrac{x}{3x+2}\\right)^5[\/latex]<\/p>\n