Question

What is the derivative of `d/dx (x+cosx ÷ tanx )`?

Answer 1

`d/dx (x+cosx ÷ tanx ) = 1 - cosx - cotxcscx`

Explanation:

`d/dx (x+cosx ÷ tanx ) = d/dx (x+cosx/tanx )`
`" "= d/dx (x)+d/dx(cosx/tanx)`
`" "= 1 + { (tanx)(-sinx) - (cosx)(sec^2x) }/ (tanx)^2`
`" "= 1 - (tanx)(sinx)/tan^2x - (cosx)(sec^2x)/(tanx)^2`
`" "= 1 - (sinx)*cosx/sinx - (cosx)(1/cos^2x)*cos^2x/sin^2x`

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

`" "= 1 - cosx - cotxcscx`