Question

How do you find the derivative of `(x+cosx)/tanx`?

Answer 1

`(df)/(dx)=(tanx-tanxsinx-xsec^2x-secx)/tan^2x`

Explanation:

We can use Quotient rule, which states if `f(x)=(g(x))/(h(x))`

then `(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2`

hence as `f(x)=(x+cosx)/tanx`

`(df)/(dx)=(tanx xx (1-sinx)-sec^2x(x+cosx))/tan^2x`

= `(tanx-tanxsinx-xsec^2x-secx)/tan^2x`