# Derivative confusion

• Apr 25th 2009, 06:29 PM
janedoe
Derivative confusion
I'm confused....is \$\displaystyle sec x^2\$ the same thing as \$\displaystyle sec^2 x\$? I was told you rewrite these as \$\displaystyle sec (x^2) and (sec x)^2\$, respectively.

Now I have a problem and I'm unsure how to do it since I don't know the above rule:

Find the derivative of f(Θ) = secΘ^2

How would you rewrite this and how would you proceed?
If I treat it as (secΘ)^2 I get: 2secΘ^2(tanΘ)

Thanks=]
• Apr 25th 2009, 06:35 PM
mr fantastic
Quote:

Originally Posted by janedoe
I'm confused....is \$\displaystyle sec x^2\$ the same thing as \$\displaystyle sec^2 x\$? I was told you rewrite these as \$\displaystyle sec (x^2) and (sec x)^2\$, respectively.

Now I have a problem and I'm unsure how to do it since I don't know the above rule:

Find the derivative of f(Θ) = secΘ^2

How would you rewrite this and how would you proceed?
If I treat it as (secΘ)^2 I get: 2secΘ^2(tanΘ)

Thanks=]

What you've posted has several interpretations. Ask your instructor which one is intended.
• Apr 25th 2009, 07:35 PM
curvature
Quote:

Originally Posted by janedoe
I'm confused....is \$\displaystyle sec x^2\$ the same thing as \$\displaystyle sec^2 x\$? I was told you rewrite these as \$\displaystyle sec (x^2) and (sec x)^2\$, respectively.

Now I have a problem and I'm unsure how to do it since I don't know the above rule:

Find the derivative of f(Θ) = secΘ^2

How would you rewrite this and how would you proceed?
If I treat it as (secΘ)^2 I get: 2secΘ^2(tanΘ)

Thanks=]

According to math notations, \$\displaystyle sec x^2=sec(x^2)\$ (squared first, sec second) and \$\displaystyle sec^2 x=(secx)^2\$ (sec first, squared second).