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Find the gradient of y with respect to x

  1. Aug 2, 2009 #1
    1. The problem statement, all variables and given/known data


    Find the gradient of y with respect to x:

    y=[tex]\frac{3\sqrt{\theta^{2}+1}}{\frac{1}{2}cos(x^{2}+2\theta)}[/tex]


    Well, I am at a complete loss where to start with this. The learning package I have in all its examples and text has no similar worded examples or utilises 2 unknowns.

    That being said, I can see that I will be utilising the product rule and quotient rule in order to differentiate to the first derivative to find the gradient.

    Can someone give me some direction in order to solve this.

    Thanks in advance.
     
  2. jcsd
  3. Aug 2, 2009 #2

    tiny-tim

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    Hi parabol! :smile:

    (i assume you know that gradient is the same as the slope of the tangent :wink:)

    Imagine that this is a surface on a 3D graph, with y up, and with x and θ horizontal (ie like x y and z, but with θ instead of z).

    Then at each point P, the gradient of y with respect to x is the slope of the surface in the x-direction (parallel to the x-axis), and the gradient of y with respect to θ is the slope of the surface in the θ-direction (parallel to the θ-axis).

    (They would normally be written ∂y/∂x and ∂y/∂θ, respectively, and are called "partial derivatives".)

    So in this case, just differentiate wrt x, keeping θ constant. :smile:
     
  4. Aug 2, 2009 #3
    Re: Gradients

    Thanks tiny-tim

    A touch of progress then.

    Utilsing the chain rule I have differentiated the denominator.

    let v = x2 + 2[tex]\theta[/tex], let u = cos v let y = 1/2u

    [tex]\frac{dv}{dx}= 2x[/tex] ...... [tex]\frac{du}{dv}= -sin.v = -sin(x^{2}+2\theta)[/tex] ...... [tex]\frac{dy}{du}= \frac{1}{2}[/tex]


    [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}[/tex]

    [tex]\frac{dy}{dx}=\frac{1}{2}.-sin(x^{2}+2\theta).2x =-x.sin(x^{2}+2\theta)[/tex]

    Following this I'd use the quotient rule to differentiate further. But treating [tex]\theta[/tex] as a constant would mean the when differentiating the numerator it would be zero surely? Or am I miss understanding this completely.
     
  5. Aug 2, 2009 #4

    tiny-tim

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    Hi parabol! :smile:

    (have a theta: θ :wink:)
    If you want to differentiate K/f(x), where K is a constant, then there's no need to use the quotient rule … it's just K time the derivative of 1/f(x). :wink:

    As to the rest, you've made it very complicated :redface:

    start again, and write 1/cos = sec. :smile:
     
  6. Aug 2, 2009 #5
    Re: Gradients

    Thanks for your help tiny-tim.

    But you have completly thrown me there. The texts I am working off are, well, limited and teach the basics. They essentially cover, sum, product, chain rule and quotient rule and then some higher derivatives, so I'm not certain what techniques I should be using not to be so completlicated.
     
  7. Aug 2, 2009 #6

    tiny-tim

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    ok … if you don't know the derivative of sec, then just use the chain rule twice, once for cosx, and once for 1/x. :wink:
     
  8. Aug 2, 2009 #7
    Re: Gradients

    that would be secxtanx.

    I do not understand the above statement. Could you elaborate?

    Thanks
     
  9. Aug 2, 2009 #8

    tiny-tim

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    Yup! :biggrin:
    If K is a constant, then the derivative of K is 0, so the product rule means that the derivative of K times 1/f(x) is 0 times 1/f(x) plus K times the derivative of 1/f(x) …

    in other words, it's just K times the derivative of 1/f(x) …

    it's so straightforward that nobody ever bothers to mention the product rule, they just take it for granted that you can take K "outside", and differentiate without it.
     
  10. Aug 2, 2009 #9
    Re: Gradients

    Things are starting to become a bit clearer, I think/hope.

    In this question then, is [tex]K=3\sqrt{\theta^{2}+1}[/tex]???

    And if it is then [tex]\frac{dy}{dx}=(3\sqrt{\theta^{2}+1})\times(-x.sin(x^{2}+2\theta))[/tex]

    If it is I can see why you siad I had complicated it. I just wouldn't know how to make it less so.
     
  11. Aug 2, 2009 #10

    tiny-tim

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    Yes, that's right :smile:

    except you've used cos() instead of 1/cos() :cry:
     
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