Thread: derivative of secx

I just wanted to make sure that I'm taking the derivative of this correctly, my problem is:
y=sec^2(x^4)
MY Answer: y'= 2sec(x^4)(sec^2 (x^4)) (tan^2 (x^4))(4x^3)

So is it right, if not please try to explain to me what I'm doing wrong because I've done this problem like 10 times and I get the same answer.

Thanks.

Originally Posted by Jasmina8

I just wanted to make sure that I'm taking the derivative of this correctly, my problem is:
y=sec^2(x^4)
MY Answer: y'= 2sec(x^4)(sec^2 (x^4)) (tan^2 (x^4))(4x^3)

So is it right, if not please try to explain to me what I'm doing wrong because I've done this problem like 10 times and I get the same answer.

Thanks.

Looks good to me. You should end up with:

8x^3 (secx^4)^2 * (tanx^4)

So I don't need the sqrd. at the second sec. and tan. ?

Originally Posted by Jasmina8

MY Answer: y'= 2sec(x^4)(sec^2 (x^4))(tan^2 (x^4))(4x^3)

sec(x^4)tan(x^4) is not squared

Originally Posted by Arturo_026

Looks good to me. You should end up with:

8x^3 (secx^4)^2 * (tanx^4)

Right. Sorry I didn't notice the square in tan