Here you will learn what is quotient rule in differentiation with examples.
Letโs begin โ
Quotient Rule in Differentiation
If f(x) and g(x) are two differentiable functions and g(x) \(\ne\) 0, then
\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) โ f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)
Example 1 : find the differentiation of \(sinx\over {x + 1}\).
Solutionย : Let f(x)ย = sinx and g(x) = x + 1
By using quotient rule in differentiation,
\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) โ f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)
\(d\over dx\) {\(sinx\over x + 1\)} = \({(x + 1) {d\over dx} (sinx) โ sinx {d\over dx} (x + 1)}\over {(x + 1)^2}\)
= \({(x + 1)(cosx) โ sinx.1}\over {(x + 1)^2}\)
= \((x+1)cosx โ sinx\over {(x + 1)^2}\)
Example 2ย : find the differentiation of \(e^x + sinx\over 1 + logx\).
Solutionย : Let f(x)ย = \(e^x + sinx\) and g(x) = 1 + logx
By using quotient rule,
\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) โ f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)
\(d\over dx\) {\(e^x + sinx\over 1 + logx\)} = \({(1 + logx) {d\over dx} (e^x + sinx) โ (e^x + sinx) {d\over dx} (1 + logx)}\over {(1 + logx)^2}\)
= \({(1 + logx) (e^x + cosx) โ (e^x + sinx) (0 + {1\over x})}\over {(1 + logx)^2}\)
= \({(1 + logx) (e^x + cosx) โ (e^x + {sinx\over x})}\over {(1 + logx)^2}\)
Example 3 : find the differentiation of \(sinx โ xcosx\over {xsinx + cosx}\).
Solutionย : Let f(x)ย = sinx โ xcosx and g(x) = xsinx + cosx
By using quotient rule,
\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) โ f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)
\(d\over dx\) {\(sinx โ xcosx\over {xsinx + cosx}\)}
= \({(xsinx + cosx) {d\over dx} (sinx โ xcosx) โ (sinx โ xcosx) {d\over dx} (xsinx + cosx)}\over {(xsinx + cosx)^2}\)
= \({(xsinx + cosx) (cosx โ cosx + xsinx) โ (sinx โ xcosx) (sinx + xcosx โ sinx)}\over {(xsinx + cosx)^2}\)
= \({(xsinx + cosx) (xsinx) โ (sinx โ xcosx) (xcosx)}\over {(xsinx + cosx)^2}\)
= \((x^2sin^2x + xsinx cosx) โ (xsinx cosx โ x^2cos^2x)\over {(xsinx + cosx)^2}\)
= \(x^2(sin^2x + cos^2x)\over {(xsinx + cosx)^2}\)
=\(x^2\over {(xsinx + cosx)^2}\)
