Question

What is the derivative of `y=tan(ln(2x+1))`?

Answer 1

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

Explanation:

Using the standard results:

`d/dx(tanx)=sec^2x` and `d/dxlnx=1/x`

Along with the chain rule then differentiating:

`y = tan(ln(2x+1))`

gives:

`dy/dx = sec^2(ln(2x+1)) * d/dx ln(2x+1)`
`\ \ \ \ \ = sec^2(ln(2x+1)) * 1/(2x+1) * d/dx(2x+1)`
`\ \ \ \ \ = sec^2(ln(2x+1)) * 1/(2x+1) * 2`
`\ \ \ \ \ = (2sec^2(ln(2x+1)))/(2x+1)`