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Math Help - derivative of secx

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Jasmina8 View Post
    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}<br />
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  3. #3
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    Looks good to me. You should end up with:

    8x^3 (secx^4)^2 * (tanx^4)
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  4. #4
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    So I don't need the sqrd. at the second sec. and tan. ?
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  5. #5
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    Quote Originally Posted by Jasmina8 View Post
    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
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  6. #6
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    Quote Originally Posted by Arturo_026 View Post
    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
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