Question

How do you find dy/dx when `y=ln(tanx)`?

Answer 1

Use the derivative of the natural logarithm and the chain rule:

`d/(dx) ln (g(x)) = 1/(g(x)) * g'(x)`

(Which can also be written: `d/(dx) ln (u) = 1/u (du)/(dx)`)

You'll also need `d/(dx)(tanx) = sec^2x`

So, for `y=ln(tanx)`, we get

`(dy)/(dx)=1/tanx sec^2x`

This can be simplified using trigonometric identities:

`(dy)/(dx)=1/tanx sec^2x = cotx * sec^2x = cosx / sinx * 1/cos^2x`

`(dy)/(dx)=1 / (sinx cosx) = cscx secx`.

OR if preferred,

`dy/dx = 2/(2sinxcosx) = 2/sin(2x) = 2csc(2x)`

Answer 2

We can also use the property that `log(a/b)=log(a)-log(b)`:

`y=ln(tanx)=ln(sinx/cosx)=ln(sinx)-ln(cosx)`

Then, to differentiate this, use the derivative of `ln(x)`, which is `1/x`, to show that `d/dxln(u)=1/u*(du)/dx`, through the chain rule. So:

`dy/dx=1/sinx(d/dxsinx)-1/cosx(d/dxcosx)`

`dy/dx=cosx/sinx-(-sinx)/cosx`

`dy/dx=(cos^2x+sin^2x)/(cosxsinx)`

`dy/dx=1/cosx(1/sinx)`

`dy/dx=secxcscx`