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Math Help - Chain Rule Problems???

  1. #1
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    Chain Rule Problems???

    Please help and explains to me how to do it...Thanks


    1. Find the equation of the tangent line to the curve
    at the point .


    2. Find the equation of the tangent line to the curve at the point .

    3. Differentiate


    4. Differentiate
    Last edited by Kayla_N; February 25th 2009 at 01:55 PM.
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  2. #2
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    1) y'=\left\{\left(3\sin x+2\cos x\right)^{-1}\right\}'=-\left(3\sin x+2\cos x\right)^{-2}\left(3\sin x+2\cos x\right)' =-\left(3\sin x+2\cos x\right)^{-2}\left(3\cos x-2\sin x\right)

    2) y'=5\sec^2 x

    3) f'(x)=\sec^2 x\left(2\sin x+3\cos x\right)+\tan x\left(2\cos x-3\sin x\right)

    4) f'(x)=2x\cos x -x^2\sin x

    For the tangent lines use \ell: \ \frac{y-y_1}{x-x_1}=\frac{dy}{dx}.
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  3. #3
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    thanks alot but how do i write the tangent line equation for number 1 and 2? y=?x+?
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  4. #4
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    Use the formula  y - b = m(x - a)
    where m = gradient, a = x co-ordinate of point on line and b = y co-ordinate of point on line
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  5. #5
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    Quote Originally Posted by Amanda H View Post
    Use the formula  y - b = m(x - a)
    where m = gradient, a = x co-ordinate of point on line and b = y co-ordinate of point on line
    Problem 1: y-5=m(x-pi/4)
    How do i find m?
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  6. #6
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    Quote Originally Posted by Kayla_N View Post
    Problem 1: y-5=m(x-pi/4)
    How do i find m?
    Problem 2 you mean ....?

    To get m, differentiate the given function and evaluate the derivative at x = pi/4 (which is why post #2 stated the derivatives for you, in case you were wondering ....)
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  7. #7
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    Oh ok..thanks

    So is it m= 5????
    Last edited by mr fantastic; February 27th 2009 at 08:42 PM. Reason: Merged posts
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  8. #8
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    Quote Originally Posted by Kayla_N View Post
    Oh ok..thanks

    So is it m= 5????
    No.

    Do you know the exact value of \cos \frac{\pi}{4} and hence the exact value of \sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} ?
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    No.

    Do you know the exact value of \cos \frac{\pi}{4} and hence the exact value of \sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} ?

    ok so 1/cos(pi/4) = 1.414. then i plug it in to tangent line equation and i got y= 1.14x-6.111 ?? is this right?
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  10. #10
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    Quote Originally Posted by Kayla_N View Post
    ok so 1/cos(pi/4) = 1.414. [snip]
    No! ..... exact value please.

    Google for the exact value of cos(pi/4) if you don't know it and can't look it up.

    Quote Originally Posted by Kayla_N View Post
    ok so 1/cos(pi/4) = 1.414. then i plug it in to tangent line equation and i got y= 1.14x-6.111 ?? is this right?
    Your gradient m is totally wrong.

    What about the squaring and multiplying by 5 to get it from the derivative?? Go back and look at what the derivative is, the one that someone went to the trouble of giving you all those posts ago.
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  11. #11
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    ok so cos (pi/4)= sqrt(2)/2. I then square it and time it by 5 and got = 5/2. so now i use the y-y1= m(x-x1) equation???
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  12. #12
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    Quote Originally Posted by Kayla_N View Post
    ok so cos (pi/4)= sqrt(2)/2. I then square it and time it by 5 and got = 5/2. so now i use the y-y1= m(x-x1) equation???
    *Sigh* Please post what the derivative is (and in doing so you will hopefully see why I'm sighing).
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  13. #13
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    omg this is so confusing..nevermind then. Thanks for your help.
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  14. #14
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    Quote Originally Posted by Kayla_N View Post
    omg this is so confusing..nevermind then. Thanks for your help.
    No. It's not confusing. You're just not concentrating. Were you not told that y' = 5 \sec^2 x ? It is very simple. You substitute x = pi/4 into this and get the exact value. That gives you the exact value of m:

    \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \Rightarrow \sec \frac{\pi}{4} = \frac{2}{\sqrt{2}} \Rightarrow \sec^2 \frac{\pi}{4} = \frac{4}{2} = 2 \Rightarrow m = 5 \sec^2 \frac{\pi}{4} = 5 (2) = 10.

    There is nothing confusing here. It simply requires a small amount of concentration and attention to detail on your part.

    Now do you see why I asked you to go back and type the derivative you were given?
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  15. #15
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    Wow...gosh I am so dumb...thanks. I should pay more attention next time and read the question carefully. Thanks again for your help..also everyone who helped me.
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