dy/dx... just making sure :)

[Lizzie]

If y=(tanx)³ dy/dx is: ?

So, we need to take the derivative of the ouside and keep the inside the same...

3(tanx)²

and then mutiply it by the derivative of the inside...

(secx)²

to get...

3[(tanx)²(secx)² ?

 

[Lizzie]

it should be 3[(tanx)²(secx)<sup2</sup>] I forgot the last bracket

 

[Lizzie]

eep! well, you know what I mean, lol

 

[opticaltempest]

Lizzie,

You are correct, \(\displaystyle 3\,{{\it tanx}}^{2}{{\it secx}}^{2}\) is the derivative of y w.r.t. x.