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Math Help - [SOLVED] more derivatives

  1. #1
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    [SOLVED] more derivatives

    Find and simplify the following derivatives:

    <br /> <br />
f(x) = cos^3 (3x^2 - 2)<br />

    <br /> <br />
f(x) = sec^2 (1 - \sqrt{4x})<br />

    <br /> <br />
f(x) = tan^3 (\sqrt{1-x^3})<br />

    thanks
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  2. #2
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    all of these are derivatives involving the use the chain rule ... let's see how you do at least one of them.
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  3. #3
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    i know they are the chain rule which is

    <br /> <br />
g'(h(x)) \cdot h'(x)<br />

    but I dont know how to do these chain rules with cos, sec and tan to exponents of more then 1
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  4. #4
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    f(x) = [\sec(1 - \sqrt{4x})]^2

     f'(x) = 2\sec(1 - \sqrt{4x}) \cdot \sec(1 - \sqrt{4x}) \cdot \tan(1 - \sqrt{4x}) \cdot \left(-\frac{1}{\sqrt{x}}\right)

    do the cosine one.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    f(x) = [\sec(1 - \sqrt{4x})]^2

     f'(x) = 2\sec(1 - \sqrt{4x}) \cdot \sec(1 - \sqrt{4x}) \cdot \tan(1 - \sqrt{4x}) \cdot \left(-\frac{1}{\sqrt{x}}\right)

    do the cosine one.

    wouldnt this one be:

    <br /> <br />
f'(x) = 2sec(1 - \sqrt{4x}) \cdot sec(1 - \sqrt{4x}) \cdot tan(1 - \sqrt{4x}) \cdot (\frac{-1}{2\sqrt{4x}})<br />

    ???
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  6. #6
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    \frac{d}{dx} (-\sqrt{4x}) =

    \frac{d}{dx} (-2\sqrt{x}) =

    -2 \cdot \frac{d}{dx} (\sqrt{x}) = -2 \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{\sqrt{x}}
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  7. #7
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    so the cos one would be

    <br /> <br />
f'(x) = 3[cos(3x^2-2)]^2 \cdot [cos(3x^2-2)] \cdot [-sin(3x^2-2)] \cdot (6x)<br />
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  8. #8
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    Quote Originally Posted by iz1hp View Post
    so the cos one would be

    <br /> <br />
f'(x) = 3[cos(3x^2-2)]^2 \cdot [cos(3x^2-2)] \cdot [-sin(3x^2-2)] \cdot (6x)<br />
    almost ...

     f'(x) = 3[cos(3x^2-2)]^2 \cdot [-sin(3x^2-2)] \cdot (6x)
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  9. #9
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    so the tan one is...

    <br /> <br />
f(x) = tan^3(\sqrt{1-x^3})<br />

    <br /> <br />
f(x) = [tan(\sqrt{1-x^3})]^3<br />

    <br /> <br />
f'(x) = 3[tan(\sqrt{1-x^3})]^2 \cdot [sec^2(\sqrt{1-x^3})] \cdot (\frac{1}{2\sqrt{1-x^3}}) \cdot (-3x^2)<br />

    ??
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  10. #10
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    ... and you said you didn't know how.



    good job. now clean up the algebra on all of them.
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  11. #11
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    ahhah thanks for explaining it to me haha :P

    if you can help me with one more, ill be set :P

    <br /> <br />
f(x) = cos(sin(cos x))<br />

    so far ive gotten to:

    <br /> <br />
f'(x) = -sin(sin(cos x)) \cdot cos(cos x) \cdot -sin x<br />

    and now im stuck
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  12. #12
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    Quote Originally Posted by iz1hp View Post
    ahhah thanks for explaining it to me haha :P

    if you can help me with one more, ill be set :P

    <br /> <br />
f(x) = cos(sin(cos x))<br />

    so far ive gotten to:

    <br /> <br />
f'(x) = -sin(sin(cos x)) \cdot cos(cos x) \cdot -sin x<br />

    and now im stuck
    Hey iz1hp.

    How about starting a new thread for this new question? It's a habit I'd like you and all users to get into. I'm going to close this thread but please make a new one!

    Jameson
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