{"id":5508,"date":"2021-09-15T17:55:48","date_gmt":"2021-09-15T12:25:48","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5508"},"modified":"2021-11-22T18:11:24","modified_gmt":"2021-11-22T12:41:24","slug":"differentiation-of-log-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-log-x\/","title":{"rendered":"Differentiation of Log x (Logarithmic Function)"},"content":{"rendered":"

Here you will learn differentiation of log x i.e logarithmic function by using first principle and its examples.<\/p>\n

Let’s begin –<\/p>\n

Differentiation of log x (Logarithmic Function) with base e and a\u00a0<\/h2>\n

(1) Differentiation of log x\u00a0 or \\(log_e x\\):<\/h3>\n
\n

The differentiation of \\(log_e x\\), x > 0 with respect to x is \\(1\\over x\\).<\/p>\n

i.e. \\(d\\over dx\\) \\(log_e x\\) = \\(1\\over x\\)<\/p>\n<\/blockquote>\n

Proof Using first Principle :<\/strong><\/h4>\n
\n

Let f(x) = \\(log_e x\\). Then, f(x + h) = \\(log_e (x + h)\\)<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(f(x + h) – f(x)\\over h\\)<\/p>\n

\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(log_e (x + h) – log_e x\\over h\\)<\/p>\n

\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(log_e (1 + h\/x)\\over h\\)<\/p>\n

Divide and multiply by x in both numerator and denominator,<\/p>\n

\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(log_e (1 + h\/x)\\over h\/x\\) \\(1\\over x\\)<\/p>\n

\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(1\\over x\\)<\/p>\n

because, [\\(lim_{h\\to 0}\\) \\(log_e (1 + x)\\over x\\) = 1]<\/p>\n

Hence, \\(d\\over dx\\) (\\(log_e x\\)) = \\(1\\over x\\)<\/p>\n<\/blockquote>\n

Example<\/strong><\/span> : What is the differentiation of log 5x ?<\/p>\n

Solution<\/strong><\/span> : Let y\u00a0 = log 5x<\/p>\n

\\(d\\over dx\\) (y) = \\(d\\over dx\\) log 5x<\/p>\n

By using chain rule,<\/p>\n

\\(d\\over dx\\) (y) = \\(1\\over 5x\\) .5 = \\(1\\over x\\)<\/p>\n

Hence, \\(d\\over dx\\) (log 5x) = \\(1\\over x\\)<\/p>\n

(2) Differentiation of \\(log_a x\\) :<\/h3>\n
\n

The differentiation of \\(log_a x\\), a > 0 , a \\(\\ne\\) 1 with respect to x is \\(1\\over x log_e a\\).<\/p>\n

i.e. \\(d\\over dx\\) \\(log_a x\\) = \\(1\\over x log_e a\\)<\/p>\n<\/blockquote>\n

Proof Using first Principle :<\/strong><\/h4>\n
\n

Let f(x) = \\(log_a x\\). Then, f(x + h) = \\(log_a (x + h)\\)<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(f(x + h) – f(x)\\over h\\)<\/p>\n

\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(log_a (x + h) – log_a x\\over h\\)<\/p>\n

\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(log_a (1 + h\/x)\\over h\\)<\/p>\n

By base changing theorem,<\/p>\n

\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(log_e (1 + h\/x)\\over (log_e a)h\\)<\/p>\n

\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(1\\over log_e a\\) \\(lim_{h\\to 0}\\) \\(log_e (1 + h\/x)\\over x(h\/x)\\)<\/p>\n

\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(1\\over xlog_e a\\)<\/p>\n

because, [\\(lim_{h\\to 0}\\) \\(log_e (1 + h\/x)\\over h\/x\\) = 1]<\/p>\n

Hence, \\(d\\over dx\\) (\\(log_a x\\)) = \\(1\\over xlog_e a\\)<\/p>\n<\/blockquote>\n

Example<\/strong><\/span> : What is the differentiation of \\(log_3 x\\) ?<\/p>\n

Solution<\/strong><\/span> : Let y\u00a0 = \\(log_3 x\\)<\/p>\n

\\(d\\over dx\\) (y) = \\(d\\over dx\\) \\(log_3 x\\)<\/p>\n

By using above formula,<\/p>\n

\\(d\\over dx\\) (y) = \\(1\\over xlog_e 3\\)<\/p>\n

Hence, \\(d\\over dx\\) (\\(log_3 x\\)) = \\(1\\over xlog_e 3\\)<\/p>\n

\n

Question for Practice<\/h3>\n

What is the Differentiation of log log x ?<\/a><\/p>\n

What is the Differentiation of x log x ?<\/a><\/p>\n

What is the differentiation of log sin x ?<\/a><\/p>\n

What is the differentiation of 1\/log x ?<\/a><\/p>\n

What is the differentiation of \\(log x^2\\) ?<\/a><\/p>\n\n\n

\n
Next – Differentiation of sin inverse x<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Differentiation of Exponential Function<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn differentiation of log x i.e logarithmic function by using first principle and its examples. Let’s begin – Differentiation of log x (Logarithmic Function) with base e and a\u00a0 (1) Differentiation of log x\u00a0 or \\(log_e x\\): The differentiation of \\(log_e x\\), x > 0 with respect to x is \\(1\\over x\\). …<\/p>\n

Differentiation of Log x (Logarithmic Function)<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[36],"tags":[287,275,286,288,289],"yoast_head":"\nDifferentiation of Log x (Logarithmic Function) - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn differentiation of log x i.e. logarithmic function with respect to x by using first principle.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/differentiation-of-log-x\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Differentiation of Log x (Logarithmic Function) - 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