Question

How do you find the derivative of `y=tan(2x)`?

Answer 1

Essentially we use the rule:

`d/dx(a* tan(k*x+h))=(a*k)/(cos^2(k*x+h))`

or

`a*k*sec^2(k*x+h)`

I'll go through the steps behind it as well:

Derivitave of `y=tan(2x)`

The first step is to separate it into `sin(x)` and `cos(x)`:

`y=sin(2x)/cos(2x)`

Then we integrate using the quotient rule:

`f(x)=g(x)/(h(x)),` `f'(x)= (g'(x)*h(x)-h'(x)*g(x))/(h(x))^2`

`y'=(cos(2x)xx2cos(2x) - sin(2x)xx-2sin(2x))/(cos^2(2x))`

`y'=(2cos^2(2x)+2sin^2(2x))/(cos^2(2x))`

`y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))`

To simplify this further we use the Pythagorean Identity:

`sin^2(x)+cos^2(x)=1`

So `2(cos^2(2x)+sin^2(2x))` becomes `2xx1=2`

Thus:

`y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))`

becomes

`y'=(2)/(cos^2(2x))`

`=2sec^2(2x)`

Answer 2

There is a simple answer to this.

`d/(dx)[tanu] = sec^2u((du)/(dx))`

Thus, with `u = 2x` (thus, `u/x = 2`), `(du)/(dx) = 2`, and

`d/(dx)[tan(2x)] = sec^2(2x)*2 = color(blue)(2sec^2(2x))`

Answer 3

We can Use Chain rule for this.

Explanation:

We have to find derivative of `color(blue)(y=tan(2x))`
Let us assume `u=2x`
Then
`color(blue)(y=tanu` & `color(blue)(u=2x`
By Chain Rule of Differenciation
`dy/dx=dy/(du)*(du)/(dx)`
`d/(dx)(tan2x)=(d/(du)(tanu))(d/dx(2x))`
`d/(dx)(tan2x)=` `sec^2u*2 ->1`
Substitute u=2x in Eqn 1
Therefore
`d/dx(tan2x)=2sec^2 2x`