Question

Solve the following problem.

Using the rule for differentiation for quotient of two functions, prove that `"d"/"dx" ("sin x"/"cos x") = sec^2"x"`. 

Answer 1

Using, `"d"/"dx"[("f"_1("x"))/("f"_2("x"))] = 1/("f"_2("x")) ("df"_1("x"))/"dx" - ("f"_1("x"))/("f"_2^2("x")) ("df"_2("x"))/"dx"`

For f₁(x) = sin x and f₂(x) = cos x

`"d"/"dx" ("sin x"/"cos x") = 1/"cos x" xx ("d"(sin "x"))/"dx" - "sin x"/cos^2"x" xx ("d"(cos "x"))/"dx"`

`= 1/"cos x" xx "cos x" - "sin x"/(cos^2"x") xx (- sin "x")`

`= 1 + (sin^2"x")/(cos^2"x") = (cos^2"x" + sin^2"x")/(cos^2"x")`

∴ `"d"/"dx"("sin x"/"cos x") = 1/(cos^2"x")  ....[sin^2"x" + cos^2"x" = 1]`

= sec²x      ....`(because 1/"cos x" = sec "x")`