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Math Help - Simple differentiation not getting lecturers answer :(

  1. #1
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    Simple differentiation not getting lecturers answer :(

    Hi he told us we can split cos^2 into cos . cos and use the chain rule. But i'm still not getting the same answer as him. Thanks

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  2. #2
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    Why don't you make some effort to learn LaTeX?
    It is so hard to read you posts.

    If y = \sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} then

    \dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} }}
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  3. #3
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    Quote Originally Posted by adam_leeds View Post
    Hi he told us we can split cos^2 into cos . cos and use the chain rule. But i'm still not getting the same answer as him. Thanks
    It's difficult to explain how to get the lecturer's answer if you don't give us that answer.
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    Quote Originally Posted by Plato View Post
    Why don't you make some effort to learn LaTeX?
    It is so hard to read you posts.

    If y = \sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} then

    \dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} }}
    Sorry, thanks.

    My lecturer got it to be negative and the two on the denominator was at the front of dlcos^2(\phi)

    You have missed off the two in the question. But i'll try and do yours then do it to mine and see if i get the same answer.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Why don't you make some effort to learn LaTeX?
    It is so hard to read you posts.

    If y = \sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} then

    \dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} }}
    I don't really understand how you do the chain rule bit

    is u = (d^2 + l^2 + dlcos(\phi))^(1/2)) and v = (cos(\phi))^(1/2)
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  6. #6
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    Quote Originally Posted by adam_leeds View Post
    I don't really understand how you do the chain rule bit

    is u = (d^2 + l^2 + dlcos(\phi))^(1/2)) and v = (cos(\phi))^(1/2)
    u=d^2+l^2+2dlcos(\phi)cos(\phi)

    I reckon the reason he said to split cos^2(\phi) into cos(\phi)*cos(\phi)

    is to conveniently apply the product rule to differentiating cos^2(\phi) while you are getting used to the chain rule.

    The chain rule is used in two ways along with the product rule in that case.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    u=d^2+l^2+2dlcos(\phi)cos(\phi)
    How do you differentiate that with the product rule when it is all square rooted?
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  8. #8
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    Quote Originally Posted by Plato View Post
    Why don't you make some effort to learn LaTeX?
    It is so hard to read you posts.

    If y = \sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} then

    \dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} }}
    Please explain how you did this
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    Quote Originally Posted by adam_leeds View Post
    Please explain how you did this
    First I did not use u-substitution. I learned basic formulas as I went along.
    It appears that you may have skipped that basic learning process.

    \left( {\sqrt {f(x)} } \right)^\prime   = \dfrac{{f'(x)}}<br />
{{2\sqrt {f(x)} }}
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  10. #10
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    Quote Originally Posted by Plato View Post
    First I did not use u-substitution. I learned basic formulas as I went along.
    It appears that you may have skipped that basic learning process.

    \left( {\sqrt {f(x)} } \right)^\prime   = \dfrac{{f'(x)}}<br />
{{2\sqrt {f(x)} }}
    Yep done it now, thank you.
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  11. #11
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    Quote Originally Posted by adam_leeds View Post
    Please explain how you did this
    If you did not want to memorise the rule given by Plato then using the chain rule

    \displaystyle y = \sqrt {d^2 + l^2 + dl\cos ^2 (\phi )}  = (d^2 + l^2 + dl\cos ^2 (\phi ))^{\frac{1}{2}}

    let \displaystyle u = d^2 + l^2 + dl\cos ^2 (\phi ) \implies y = u^{\frac{1}{2}}

    Finally \displaystyle \frac{dy}{d\phi} = \frac{du}{d\phi}\times \frac{dy}{du}
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  12. #12
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    Quote Originally Posted by adam_leeds View Post
    How do you differentiate that with the product rule when it is all square rooted?
    Following pickslides post above,

    then you can use the product rule on \displaystyle\frac{du}{d\phi}

    by writing cos^2\phi=cos\phi*cos\phi

    though you don't need to.
    You could also apply the chain rule to that.

    Better again, learn Plato's method.
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  13. #13
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    Quote Originally Posted by Plato View Post
    Why don't you make some effort to learn LaTeX?
    It is so hard to read you posts.

    If y = \sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} then

    \dfrac{{dy}}{{d\phi }} = \dfrac{{2dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2  + l^2  + dl\cos ^2 (\phi )} }}

    One thing that's bothering adam_leeds is that Plato has a minor typo in his expressions, but they are consistent with each other. (Plato did mention that using LaTeX would make the problem more readable.)

    They should read:

    "If y = \sqrt {d^2  + l^2  + 2dl\cos ^2 (\phi )} then

    \dfrac{{dy}}{{d\phi }} = \dfrac{{4dl\cos (\phi )\left( { - \sin (\phi )} \right)}}{{2\sqrt {d^2  + l^2  + 2dl\cos ^2 (\phi )} }}"


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