{"id":5123,"date":"2021-09-08T18:18:43","date_gmt":"2021-09-08T12:48:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5123"},"modified":"2021-11-22T22:09:11","modified_gmt":"2021-11-22T16:39:11","slug":"product-rule-in-differentiation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/product-rule-in-differentiation\/","title":{"rendered":"Product Rule in Differentiation with Examples"},"content":{"rendered":"
Here you will learn what is product rule in differentiation with examples.<\/p>\n
Let’s begin –<\/p>\n
If f(x) and g(x) are differentiable functions, then f(x)g(x) is also differentiable function such that<\/p>\n
\n\\(d\\over dx\\) {f(x) g(x)} = \\(d\\over dx\\) (f(x)) g(x) + f(x). \\(d\\over dx\\) (g(x))<\/strong><\/p>\n<\/blockquote>\n
If f(x), g(x) and h(x) are differentiable functions, then<\/p>\n
\n\\(d\\over dx\\) (f(x) g(x) h(x)) = \\(d\\over dx\\) (f(x)) g(x) h(x) + f(x). \\(d\\over dx\\) (g(x)) h(x) + f(x) g(x) \\(d\\over dx\\) (h(x)) <\/strong> <\/p>\n<\/blockquote>\n
Example 1<\/strong><\/span> : find the differentiation of sinx cosx.<\/p>\n
Solution<\/strong> <\/span>: Let sinx = f(x) and g(x) = cosx<\/p>\n
Then, by using product rule in differentiation,<\/p>\n
\\(d\\over dx\\) {f(x) g(x)} = \\(d\\over dx\\) (f(x)) g(x) + f(x). \\(d\\over dx\\) (g(x))<\/p>\n
\\(d\\over dx\\) [sinx.cosx] = \\(d\\over dx\\) (sinx) cosx + sinx. \\(d\\over dx\\) (cosx)<\/p>\n
= cosx cosx + sinx (-sinx)<\/p>\n
= \\(cos^2x\\) – \\(sin^2x\\)<\/p>\n
= cos2x<\/p>\n
Example 2<\/strong><\/span> : find the differentiation of x sinx.<\/p>\n
Solution<\/strong> <\/span>: Let x = f(x) and g(x) = sinx<\/p>\n
Then, by using product rule in differentiation,<\/p>\n
\\(d\\over dx\\) {f(x) g(x)} = \\(d\\over dx\\) (f(x)) g(x) + f(x). \\(d\\over dx\\) (g(x))<\/p>\n
\\(d\\over dx\\) [x.sinx] = \\(d\\over dx\\) (x) sinx + x. \\(d\\over dx\\) (sinx)<\/p>\n
= 1.sinx + x.(cosx)<\/p>\n
= sinx + x cosx<\/p>\n
Example 3<\/strong><\/span> : find the differentiation of \\(e^x log\\sqrt{x} tanx\\).<\/p>\n
Solution<\/strong> <\/span>: Let \\(e^x\\) = f(x) , g(x) = \\(log\\sqrt{x}\\) and h(x) = tanx<\/p>\n
Then, by using product rule,<\/p>\n
\\(d\\over dx\\) {f(x) g(x) h(x)} = \\(d\\over dx\\) (f(x)) g(x) h(x) + f(x). \\(d\\over dx\\) (g(x)) h(x) + f(x) g(x) \\(d\\over dx\\) (h(x))<\/p>\n
\\(d\\over dx\\) [ \\(e^x log\\sqrt{x} tanx\\)] = \\(d\\over dx\\) [ \\(e^x \\times {1\\over 2} logx \\times tanx\\)]<\/p>\n
= \\(1\\over 2\\) \\(d\\over dx\\) [ \\(e^x logx tanx\\)]<\/p>\n
= \\(1\\over 2\\) [{\\(d\\over dx\\) (\\(e^x\\))} logx tanx + \\(e^x\\). {\\(d\\over dx\\) (logx)} tanx + \\(e^x\\) logx {\\(d\\over dx\\) (tanx)}]<\/p>\n
= \\(1\\over 2\\) { \\(e^x\\) logx tanx + \\(e^x\\). {\\(1\\over x\\)} tanx + \\(e^x\\) logx \\(sec^2x\\)}<\/p>\n
= \\(1\\over 2\\) \\(e^x\\) { logx tanx + \\(tanx\\over x\\) + logx \\(sec^2x\\)}<\/p>\n\n\n