{"id":5125,"date":"2021-09-08T18:20:32","date_gmt":"2021-09-08T12:50:32","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5125"},"modified":"2021-11-22T22:06:48","modified_gmt":"2021-11-22T16:36:48","slug":"quotient-rule-in-differentiation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/quotient-rule-in-differentiation\/","title":{"rendered":"Quotient Rule in Differentiation with Examples"},"content":{"rendered":"
Here you will learn what is quotient rule in differentiation with examples.<\/p>\n
Let’s begin –<\/p>\n
If f(x) and g(x) are two differentiable functions and g(x) \\(\\ne\\) 0, then<\/p>\n
\n\\(d\\over dx\\) {\\(f(x)\\over g(x)\\)} = \\({g(x) {d\\over dx} (f(x)) – f(x) {d\\over dx} (g(x))}\\over {(g(x))^2}\\)<\/p>\n<\/blockquote>\n
Example 1<\/span><\/strong> : find the differentiation of \\(sinx\\over {x + 1}\\).<\/p>\n
Solution <\/span><\/strong>: Let f(x) = sinx and g(x) = x + 1<\/p>\n
By using quotient rule in differentiation,<\/p>\n
\\(d\\over dx\\) {\\(f(x)\\over g(x)\\)} = \\({g(x) {d\\over dx} (f(x)) – f(x) {d\\over dx} (g(x))}\\over {(g(x))^2}\\)<\/p>\n
\\(d\\over dx\\) {\\(sinx\\over x + 1\\)} = \\({(x + 1) {d\\over dx} (sinx) – sinx {d\\over dx} (x + 1)}\\over {(x + 1)^2}\\)<\/p>\n
= \\({(x + 1)(cosx) – sinx.1}\\over {(x + 1)^2}\\)<\/p>\n
= \\((x+1)cosx – sinx\\over {(x + 1)^2}\\)<\/p>\n
Example 2<\/span><\/strong> : find the differentiation of \\(e^x + sinx\\over 1 + logx\\).<\/p>\n
Solution <\/span><\/strong>: Let f(x) = \\(e^x + sinx\\) and g(x) = 1 + logx<\/p>\n
By using quotient rule,<\/p>\n
\\(d\\over dx\\) {\\(f(x)\\over g(x)\\)} = \\({g(x) {d\\over dx} (f(x)) – f(x) {d\\over dx} (g(x))}\\over {(g(x))^2}\\)<\/p>\n
\\(d\\over dx\\) {\\(e^x + sinx\\over 1 + logx\\)} = \\({(1 + logx) {d\\over dx} (e^x + sinx) – (e^x + sinx) {d\\over dx} (1 + logx)}\\over {(1 + logx)^2}\\)<\/p>\n
= \\({(1 + logx) (e^x + cosx) – (e^x + sinx) (0 + {1\\over x})}\\over {(1 + logx)^2}\\)<\/p>\n
= \\({(1 + logx) (e^x + cosx) – (e^x + {sinx\\over x})}\\over {(1 + logx)^2}\\)<\/p>\n
Example 3<\/span><\/strong> : find the differentiation of \\(sinx – xcosx\\over {xsinx + cosx}\\).<\/p>\n
Solution <\/span><\/strong>: Let f(x) = sinx – xcosx and g(x) = xsinx + cosx<\/p>\n
By using quotient rule,<\/p>\n
\\(d\\over dx\\) {\\(f(x)\\over g(x)\\)} = \\({g(x) {d\\over dx} (f(x)) – f(x) {d\\over dx} (g(x))}\\over {(g(x))^2}\\)<\/p>\n
\\(d\\over dx\\) {\\(sinx – xcosx\\over {xsinx + cosx}\\)}<\/p>\n
= \\({(xsinx + cosx) {d\\over dx} (sinx – xcosx) – (sinx – xcosx) {d\\over dx} (xsinx + cosx)}\\over {(xsinx + cosx)^2}\\)<\/p>\n
= \\({(xsinx + cosx) (cosx – cosx + xsinx) – (sinx – xcosx) (sinx + xcosx – sinx)}\\over {(xsinx + cosx)^2}\\)<\/p>\n
= \\({(xsinx + cosx) (xsinx) – (sinx – xcosx) (xcosx)}\\over {(xsinx + cosx)^2}\\)<\/p>\n
= \\((x^2sin^2x + xsinx cosx) – (xsinx cosx – x^2cos^2x)\\over {(xsinx + cosx)^2}\\)<\/p>\n
= \\(x^2(sin^2x + cos^2x)\\over {(xsinx + cosx)^2}\\)<\/p>\n
=\\(x^2\\over {(xsinx + cosx)^2}\\)<\/p>\n\n\n