{"id":5506,"date":"2021-09-15T17:03:32","date_gmt":"2021-09-15T11:33:32","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5506"},"modified":"2021-11-22T19:30:50","modified_gmt":"2021-11-22T14:00:50","slug":"differentiation-of-exponential-function","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-exponential-function\/","title":{"rendered":"Differentiation of Exponential Function"},"content":{"rendered":"

Here you will learn differentiation of exponential function by using first principle and its examples.<\/p>\n

Let’s begin –<\/p>\n

Differentiation of Exponential Function<\/h2>\n

(1) Differentiation of \\(e^x\\) :<\/h3>\n
\n

The differentiation of \\(e^x\\) with respect to x is \\(e^x\\).<\/p>\n

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

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

Let f(x) = \\(e^x\\). Then, f(x + h) = \\(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}\\) \\(e^{x + h} – e^x\\over h\\)<\/p>\n

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

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

\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(e^x\\)\u00a0 \\(lim_{h\\to 0}\\) (\\(e^h – 1\\over h\\))<\/p>\n

because, [\\(lim_{h\\to 0}\\)(\\(e^h – 1\\over h\\)) = 1]<\/p>\n

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

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

Example<\/span><\/strong> : What is the differentiation of \\(e^{2x}\\) ?<\/p>\n

Solution<\/strong><\/span> : Let y\u00a0 = \\(e^{2x}\\)<\/p>\n

\\(d\\over dx\\) (y) = \\(d\\over dx\\) \\(e^{2x}\\)<\/p>\n

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

\\(d\\over dx\\) (y) = 2\\(e^{2x}\\)<\/p>\n

Hence, \\(d\\over dx\\) (\\(e^{2x}\\)) = 2\\(e^{2x}\\)<\/p>\n

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

The differentiation of \\(a^x\\) with respect to x is \\(a^x log_e a\\).<\/p>\n

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

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

Let f(x) = \\(a^x\\). Then, f(x + h) = \\(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}\\) \\(a^{x + h} – a^x\\over h\\)<\/p>\n

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

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

\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(a^x\\)\u00a0 \\(lim_{h\\to 0}\\) (\\(a^h – 1\\over h\\))<\/p>\n

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

\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(a^x\\) \\(\\times\\) \\(log_e a\\) = \\(a^x\\) \\(log_e a\\)<\/p>\n

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

Example<\/span><\/strong> : What is the differentiation of \\(5^{x}\\) ?<\/p>\n

Solution<\/strong><\/span> : Let y\u00a0 = \\(5^{x}\\)<\/p>\n

\\(d\\over dx\\) (y) = \\(d\\over dx\\) \\(5^{x}\\)<\/p>\n

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

Hence, \\(d\\over dx\\) (\\(5^{x}\\)) = \\(5^x log_e 5\\)<\/p>\n

\n

Related Questions<\/h3>\n

What is the differentiation of \\(e^{sinx}\\) ?<\/a><\/p>\n

What is the integration of \\(e^x\\) ?<\/a><\/p>\n\n\n

\n
Next – Differentiation of Log x (Logarithmic Function)<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Differentiation of cosecx<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn differentiation of exponential function by using first principle and its examples. Let’s begin – Differentiation of Exponential Function (1) Differentiation of \\(e^x\\) : The differentiation of \\(e^x\\) with respect to x is \\(e^x\\). i.e. \\(d\\over dx\\) \\(e^x\\) = \\(e^x\\) Proof Using first Principle : Let f(x) = \\(e^x\\). Then, f(x + …<\/p>\n

Differentiation of Exponential 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":[275,292,291,290],"yoast_head":"\nDifferentiation of Exponential Function - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn differentiation of exponential function by using first principle and its examples.\" \/>\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-exponential-function\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Differentiation of Exponential Function - 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