# Prove: is y = tan(2x), than y" = 4y . y' ? Thank you!t

Jun 10, 2018

We can compute the first derivative as being

$y ' = 2 {\sec}^{2} \left(2 x\right)$

Now

$4 y y ' = 4 \left(\tan \left(2 x\right)\right) \left(2 {\sec}^{2} \left(2 x\right)\right) = 8 \tan \left(2 x\right) {\sec}^{2} \left(2 x\right)$

Therefore this is saying that

$y ' ' = 8 \tan \left(2 x\right) {\sec}^{2} \left(2 x\right)$

Let's verify this by differentiating the first derivative.

We can rewrite as

$y ' = 2 {\left(\sec \left(2 x\right)\right)}^{2}$

y'' = 2(2sec(2x)sec(2x)tan(2x)(2)

$y ' ' = 8 {\sec}^{2} \left(2 x\right) \tan \left(2 x\right)$

As required. Therefore we've proved the statement.

Hopefully this helps!