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Math Help - New Calculus Problem

  1. #1
    Newbie
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    Sorry to ressurect this, but I have another assignment on similar subject matter. I figured it would be better to ask again her, than make a new thread.

    Both of these types I remember covering in class, but can't seem to find them in my notes.

    e^{2x^3}
    That is e to the power of 2 times x to the power of 3.

    Do I double up the chain rule here or something?

    Also
    <br />
sin(tan(x))

    would that just become
    <br />
-cos(sec^2(x))

    Or is there more to it?
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  2. #2
    Grand Panjandrum
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    Don't tag a new question onto the end of an existing thread, it makes it
    difficult to identify that a new question has been asked.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Sucker Punch View Post
    Sorry to ressurect this, but I have another assignment on similar subject matter. I figured it would be better to ask again her, than make a new thread.

    Both of these types I remember covering in class, but can't seem to find them in my notes.

    e^{2x^3}
    That is e to the power of 2 times x to the power of 3.

    Do I double up the chain rule here or something?
    Chain rule:

    <br />
\frac{d}{dx}f(g(x))=g'(x) f'(g(x))<br />

    In this case g(x)=2x^3 and f(x)=e^x, so

    g'(x)=6x^2

    and

    f'(x)=e^x

    So:

    <br />
\frac{d}{dx}e^{2x^3}=6x^2\ e^{2x^2}<br />

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Sucker Punch View Post
    Also
    <br />
sin(tan(x))

    would that just become
    <br />
-cos(sec^2(x))

    Or is there more to it?
    Chain rule:

    <br />
\frac{d}{dx}[\sin(\tan(x))]=\left\{ \frac{d}{dx}[\tan(x)]\right\}\ \cos(\tan(x)) = sec^2(x)\ \cos(\tan(x))<br />

    RonL
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