Quotient Rule in Differentiation with Examples

Here you will learn what is quotient rule in differentiation with examples.

Let’s begin –

Quotient Rule in Differentiation

If f(x) and g(x) are two differentiable functions and g(x) \(\ne\) 0, then

\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) – f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)

Example 1 : find the differentiation of \(sinx\over {x + 1}\).

Solution : Let f(x)  = sinx and g(x) = x + 1

By using quotient rule in differentiation,

\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) – f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)

\(d\over dx\) {\(sinx\over x + 1\)} = \({(x + 1) {d\over dx} (sinx) – sinx {d\over dx} (x + 1)}\over {(x + 1)^2}\)

= \({(x + 1)(cosx) – sinx.1}\over {(x + 1)^2}\)

= \((x+1)cosx – sinx\over {(x + 1)^2}\)

Example 2 : find the differentiation of \(e^x + sinx\over 1 + logx\).

Solution : Let f(x)  = \(e^x + sinx\) and g(x) = 1 + logx

By using quotient rule,

\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) – f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)

\(d\over dx\) {\(e^x + sinx\over 1 + logx\)} = \({(1 + logx) {d\over dx} (e^x + sinx) – (e^x + sinx) {d\over dx} (1 + logx)}\over {(1 + logx)^2}\)

= \({(1 + logx) (e^x + cosx) – (e^x + sinx) (0 + {1\over x})}\over {(1 + logx)^2}\)

= \({(1 + logx) (e^x + cosx) – (e^x + {sinx\over x})}\over {(1 + logx)^2}\)

Example 3 : find the differentiation of \(sinx – xcosx\over {xsinx + cosx}\).

Solution : Let f(x)  = sinx – xcosx and g(x) = xsinx + cosx

By using quotient rule,

\(d\over dx\) {\(f(x)\over g(x)\)} = \({g(x) {d\over dx} (f(x)) – f(x) {d\over dx} (g(x))}\over {(g(x))^2}\)

\(d\over dx\) {\(sinx – xcosx\over {xsinx + cosx}\)}

= \({(xsinx + cosx) {d\over dx} (sinx – xcosx) – (sinx – xcosx) {d\over dx} (xsinx + cosx)}\over {(xsinx + cosx)^2}\)

= \({(xsinx + cosx) (cosx – cosx + xsinx) – (sinx – xcosx) (sinx + xcosx – sinx)}\over {(xsinx + cosx)^2}\)

= \({(xsinx + cosx) (xsinx) – (sinx – xcosx) (xcosx)}\over {(xsinx + cosx)^2}\)

= \((x^2sin^2x + xsinx cosx) – (xsinx cosx – x^2cos^2x)\over {(xsinx + cosx)^2}\)

= \(x^2(sin^2x + cos^2x)\over {(xsinx + cosx)^2}\)

=\(x^2\over {(xsinx + cosx)^2}\)

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