# derivative of secx

• Oct 23rd 2009, 05:01 PM
Jasmina8
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.
• Oct 23rd 2009, 05:06 PM
skeeter
Quote:

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.

$\frac{d}{dx}[\sec^2{u}] = 2\sec{u} \cdot \sec{u}\tan{u} \cdot \frac{du}{dx}
$
• Oct 23rd 2009, 05:07 PM
Arturo_026
Looks good to me. You should end up with:

8x^3 (secx^4)^2 * (tanx^4)
• Oct 23rd 2009, 05:07 PM
Jasmina8
So I don't need the sqrd. at the second sec. and tan. ?
• Oct 23rd 2009, 05:12 PM
skeeter
Quote:

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
• Oct 23rd 2009, 05:14 PM
Arturo_026
Quote:

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