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Thread: derivatives

  1. #1
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    derivatives

    the derivative of

    y = e^secx

    is

    y = e^secx correct?
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  2. #2
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    Nope, not correct. You forgot to apply the chain rule.

    Recall that:
    $\displaystyle \frac{d}{dx} \sec{x} = \sec{x}\tan{x}$
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  3. #3
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    Quote Originally Posted by jvignacio View Post
    the derivative of

    y = e^secx

    is

    y = e^secx correct?
    Incorrect.

    We know that:

    $\displaystyle \frac{d}{dx}e^x=e^x$

    But this time our exponent is $\displaystyle \sec{x}$, not $\displaystyle x$. So what we must do is substitute. Let $\displaystyle u=\sec{x}$. Then:

    $\displaystyle e^{\sec{x}}=e^u$

    $\displaystyle \frac{d}{du}e^u=e^u=e^{\sec{x}}$

    $\displaystyle \frac{dy}{du}=e^{\sec{x}}$

    But we can't stop there! What we have taken is $\displaystyle \frac{dy}{du}$. To get $\displaystyle \frac{dy}{dx}$, we have to multiply by $\displaystyle \frac{du}{dx}$, thusly:

    $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

    So, what is $\displaystyle \frac{du}{dx}$? Well, that's simple enough to find:

    $\displaystyle u=\sec{x}$

    $\displaystyle \frac{du}{dx}=\frac{d}{dx}\sec{x}=\sec{x}\tan{x}$

    $\displaystyle \frac{du}{dx}=\sec{x}\tan{x}$

    So:

    $\displaystyle \frac{dy}{du}\frac{du}{dx}=e^{\sec{x}}\sec{x}\tan{ x}$

    And, so, our answer is:

    $\displaystyle \frac{dy}{dx}=e^{\sec{x}}\sec{x}\tan{x}$

    In the future, you can skip writing down everything and just use this chain rule in your head.
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  4. #4
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    Quote Originally Posted by Chop Suey View Post
    Nope, not correct. You forgot to apply the chain rule.

    Recall that:
    $\displaystyle \frac{d}{dx} \sec{x} = \sec{x}\tan{x}$
    y' = e^secx(secxtanx)

    better?
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  5. #5
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    Quote Originally Posted by hatsoff View Post
    Incorrect.

    We know that:

    $\displaystyle \frac{d}{dx}e^x=e^x$

    But this time our exponent is $\displaystyle \sec{x}$, not $\displaystyle x$. So what we must do is substitute. Let $\displaystyle u=\sec{x}$. Then:

    $\displaystyle e^{\sec{x}}=e^u$

    $\displaystyle \frac{d}{du}e^u=e^u=e^{\sec{x}}$

    $\displaystyle \frac{dy}{du}=e^{\sec{x}}$

    But we can't stop there! What we have taken is $\displaystyle \frac{dy}{du}$. To get $\displaystyle \frac{dy}{dx}$, we have to multiply by $\displaystyle \frac{du}{dx}$, thusly:

    $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

    So, what is $\displaystyle \frac{du}{dx}$? Well, that's simple enough to find:

    $\displaystyle u=\sec{x}$

    $\displaystyle \frac{du}{dx}=\frac{d}{dx}\sec{x}=\sec{x}\tan{x}$

    $\displaystyle \frac{du}{dx}=\sec{x}\tan{x}$

    So:

    $\displaystyle \frac{dy}{du}\frac{du}{dx}=e^{\sec{x}}\sec{x}\tan{ x}$

    And, so, our answer is:

    $\displaystyle \frac{dy}{dx}=e^{\sec{x}}\sec{x}\tan{x}$

    In the future, you can skip writing down everything and just use this chain rule in your head.
    thank u for that explanation ! much appreciated!
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