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Dy/dx if e^2x = sin(x+3y)

  1. Nov 20, 2005 #1
    I need help with this problem.

    Find dy/dx if e^2x = sin(x+3y)

    Any help would be great!

    I began by taking the natural log of both sides...which may not be correct??

    Thanks!
     
  2. jcsd
  3. Nov 20, 2005 #2

    mathwonk

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    then what?
     
  4. Nov 20, 2005 #3

    HallsofIvy

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    That's certainly one way to do it: you get 2x= ln(sin(x+3y)).
    Of course, you will have to use the chain rule on the right side when you differentiate now.
    Couldn't you have used the chain rule on both sides of the equation the way it was written? What is [itex]\frac{de^u}{du}[/itex]? What is
    [itex]\frac{de^u}{du}\frac{du}{dx}[/itex] if u= 2x?
     
  5. Nov 20, 2005 #4
    So [tex] e^{2x} = \sin(x+3y) [/tex] When you take the natural log of both sides you get: [tex] \ln e^{2x} = \ln\sin(x+3y) [/tex] or [tex] 2x = \ln\sin(x+3y) [/tex]. Now use implicit differentiation to find [tex] \frac{dy}{dx} [/tex]
     
  6. Nov 20, 2005 #5
    Ok, right.

    So

    2x = ln (sin(x+3y))

    Then if you take the derivative...

    2 = (1/sin(x+3y)) * cos (x+3y) * 3 * dy/dx

    So dy/dx = 2(sin(x+3y))/ 3(cos(x+3y)) ?

    Is that right?
     
  7. Nov 20, 2005 #6
    yes that is correct.
     
  8. Nov 20, 2005 #7
    Thanks so much! Does that simplify any farther?
     
  9. Nov 20, 2005 #8
    I guess you could write it as [tex] \frac{2\tan(x+3y)}{3} [/tex]
     
  10. Nov 20, 2005 #9
    Ok good! That is what I thought.

    You have been so helpful! Thanks!

    I have a couple more [shorter] questions, however don't feel obliged to respond if you don't want to...

    1. Find dy/dx of y = x^5 * 7^x

    2. Find dy/dx of y = (1/2) ^ x

    3. Find dy/dx of y = (cosx) ^ x
     
  11. Nov 20, 2005 #10
    1. Use the chain rule. [tex] y = x^{5}7^{x} [/tex]. [tex] y' = x^{5}(7^{x}\ln 7) + 7^{x}(5x^{4}) [/tex]
    2. [tex] (0.5)^{x} [/tex] is just [tex] (0.5)^{x}\ln 0.5 [/tex]
    3. I'll let you do this one

    In general, [tex] \frac{d}{dx} a^{u} = a^{u}\ln a \frac{du}{dx} [/tex]
     
  12. Nov 20, 2005 #11
    Right, right. That rule would make those very easy! Thanks bud! :tongue:

    I am still having trouble with the cosine one however. I apparently need to use multiple steps to find the derivative.
     
    Last edited: Nov 20, 2005
  13. Nov 21, 2005 #12

    HallsofIvy

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    Mistype in 2: [tex](0.5)^x[/tex] is just [tex]e^x\ln 0.5[/tex].

    By the way, I don't see anything gained by taking the logarithm in the first problem you posted. It is just as easy to differentiate both sides of
    e2x= sin(x+ 3y) the way it stands:
    2e2x= cos(x+ 3y)(1+ y') so
    [tex]1+ y'= \frac{2e^{2x}}{cos(x+3y}}[/tex]
    and
    [tex]y'= \frac{2e^{2x}}{cos(x+3y}}- 1[/tex]
     
  14. Nov 21, 2005 #13

    VietDao29

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    Nope, that's {B]not[/B] correct!
    The above line should read:
    [tex]2 = \frac{\cos (x + 3y)}{\sin (x + 3y)} \left( 1 + 3 \frac{dy}{dx} \right)[/tex]
    You forgot the (1 + 3 dy / dx)
    ???
    How on Earth can that be called correct? :confused:
    That's also a mistype.
    0.5x = ex ln(0.5)
    By the way, there's a typo there:
    e2x= sin(x+ 3y)
    Differentiate both sides with respect to x gives:
    2e2x= cos(x+ 3y)(1+ 3y')



    ---------------------
    To find dy / dx of y = cosxx. Just do the same:
    cosxx = ex ln(cos(x))
    ---------------------
    Or, you can do it a bit differently:
    y = cosxx
    Take the logarithm of both sides gives:
    ln y = x ln(cos x)
    Then you can differentiate both sides with respect to x, and see what you get.
     
    Last edited: Nov 21, 2005
  15. Nov 21, 2005 #14
    sorry for the typos

    Thanks HallsofIvy and VietDao for correcting me.
     
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