Question

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

Answer 1

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

Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If `y=f(x)` then `f'(x)=dy/dx=dy/(du)(du)/dx`

I was taught to remember that the differential can be treated like a fraction and that the "`dx`'s" of a common variable will "cancel" (It is important to realise that `dy/dx` isn't a fraction but an operator that acts on a function, there is no such thing as "`dx`" or "`dy`" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

`dy/dx = dy/(dv)(dv)/(du)(du)/dx` etc, or `(dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx)`

So with `y = tan(sqrt(2x+1)`, Then:

`{ ("Let "u=2x+1, => , (du)/dx=2), ("And "v=sqrtu, =>, (dv)/(du)=1/2u^(-1/2)=1/(2sqrtu)),("Then "y=tanv, =>, (dy)/(dv)=sec^2v ) :}`

Using `dy/dx=(dy/(dv))((dv)/(du))((du)/dx)` we get:

`dy/dx = (sec^2v)(1/(2sqrtu))(2)`
`:. dy/dx = (sec^2(sqrtu))(1/(sqrtu))`
`:. dy/dx = (sec^2(sqrt(2x+1))) /(sqrt(2x+1))`