{"id":265,"date":"2023-09-20T22:48:40","date_gmt":"2023-09-20T22:48:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivative-of-the-logarithmic-function\/"},"modified":"2024-08-05T12:57:19","modified_gmt":"2024-08-05T12:57:19","slug":"derivatives-of-exponential-and-logarithmic-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-exponential-and-logarithmic-functions-learn-it-2\/","title":{"raw":"Derivatives of Exponential and Logarithmic Functions: Learn It 2","rendered":"Derivatives of Exponential and Logarithmic Functions: Learn It 2"},"content":{"raw":"
Derivative of the Logarithmic Function<\/h2>\r\n
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.<\/p>\r\n\r\n
derivative of the natural logarithmic function<\/h3>\r\n
If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then<\/p>\r\n
More generally, let [latex]g(x)[\/latex] be a differentiable function. For all values of [latex]x[\/latex] for which [latex]g^{\\prime}(x)>0[\/latex], the derivative of [latex]h(x)=\\ln(g(x))[\/latex] is given by<\/p>\r\n
If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then [latex]e^y=x[\/latex]. Differentiating both sides of this equation results in the equation<\/p>\r\n
[latex]e^y\\frac{dy}{dx}=1[\/latex]<\/div>\r\n
<\/p>\r\n
Solving for [latex]\\frac{dy}{dx}[\/latex] yields<\/p>\r\n
We may also derive this result by applying the inverse function theorem, as follows. Since [latex]y=g(x)=\\ln x[\/latex] is the inverse of [latex]f(x)=e^x[\/latex], by applying the inverse function theorem we have<\/p>\r\n
The graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in Figure 3.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] Figure 3.\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].[\/caption]\r\n\r\n\r\n
Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex]<\/p>\r\n
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.<\/p>\r\n
[latex]\\begin{array}{lllll} f(x) & = \\ln(\\frac{x^2 \\sin x}{2x+1})=2\\ln x+\\ln(\\sin x)-\\ln(2x+1) & & & \\text{Apply properties of logarithms.} \\\\ f^{\\prime}(x) & = \\frac{2}{x} + \\frac{\\cos x}{\\sin x} -\\frac{2}{2x+1} & & & \\text{Apply sum rule and} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ & = \\frac{2}{x} + \\cot x - \\frac{2}{2x+1} & & & \\text{Simplify using the quotient identity for cotangent.} \\end{array}[\/latex]<\/div>\r\n\r\n[\/hidden-answer]<\/section>\r\n
Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of [latex]y=\\log_b x[\/latex] and [latex]y=b^x[\/latex] for [latex]b>0, \\, b\\ne 1[\/latex].<\/p>\r\n\r\n
derivatives of general exponential and logarithmic functions<\/h3>\r\n
Let [latex]b>0, \\, b\\ne 1[\/latex], and let [latex]g(x)[\/latex] be a differentiable function.<\/p>\r\n
If [latex]y=\\log_b x[\/latex], then [latex]b^y=x[\/latex]. It follows that [latex]\\ln(b^y)=\\ln x[\/latex]. Thus [latex]y \\ln b = \\ln x[\/latex]. Solving for [latex]y[\/latex], we have [latex]y=\\frac{\\ln x}{\\ln b}[\/latex]. Differentiating and keeping in mind that [latex]\\ln b[\/latex] is a constant, we see that<\/p>\r\n
The derivative from above now follows from the chain rule.<\/p>\r\n
If [latex]y=b^x[\/latex], then [latex]\\ln y=x \\ln b[\/latex]. Using implicit differentiation, again keeping in mind that [latex]\\ln b[\/latex] is constant, it follows that [latex]\\frac{1}{y}\\frac{dy}{dx}=\\text{ln}b.[\/latex] Solving for [latex]\\frac{dy}{dx}[\/latex] and substituting [latex]y=b^x[\/latex], we see that<\/p>\r\n
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.<\/p>\n\n
derivative of the natural logarithmic function<\/h3>\n
If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then<\/p>\n
More generally, let [latex]g(x)[\/latex] be a differentiable function. For all values of [latex]x[\/latex] for which [latex]g^{\\prime}(x)>0[\/latex], the derivative of [latex]h(x)=\\ln(g(x))[\/latex] is given by<\/p>\n
If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then [latex]e^y=x[\/latex]. Differentiating both sides of this equation results in the equation<\/p>\n
[latex]e^y\\frac{dy}{dx}=1[\/latex]<\/div>\n
<\/p>\n
Solving for [latex]\\frac{dy}{dx}[\/latex] yields<\/p>\n
We may also derive this result by applying the inverse function theorem, as follows. Since [latex]y=g(x)=\\ln x[\/latex] is the inverse of [latex]f(x)=e^x[\/latex], by applying the inverse function theorem we have<\/p>\n
The graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in Figure 3.<\/p>\nFigure 3.\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].<\/figcaption><\/figure>\n\n
Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex]<\/p>\n
![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205536\/CNX_Calc_Figure_03_09_003.jpg\")
Figure 3.\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].[\/caption]\r\n\r\n