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    Completely confused, any help appreciated
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    (Original post by poppydoodle)
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    Completely confused, any help appreciated
    To differentiate use the product rule

    \dfrac{d}{dx} (uv) = v \dfrac{du}{dx} + u \dfrac{dv}{dx}

    where u = cos x and v = e^2x.

    For (a) set dy/dx =0

    For (b), calculate the gradient at x=0 and the y-value at x=0. Then find the equation of the tangent
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    (Original post by poppydoodle)

    Completely confused, any help appreciated
    Have you covered the product rule?
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    dy/dx = cos(x) 2e^2x - e^2x sin(x)

    Is this right?
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    (Original post by poppydoodle)
    dy/dx = cos(x) 2e^2x - e^2x sin(x)

    Is this right?
    Yes.

    Do you know how to carry on with the question?
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    (Original post by RDKGames)
    Yes.

    Do you know how to carry on with the question?
    I know you set it equal to 0 but I don’t know how to solve it
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    dy/dx = cos(x) 2e^2x - e^2x sin(x)

    Factor e^2x.
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    (Original post by poppydoodle)
    I know you set it equal to 0 but I don’t know how to solve it
    You do not need to solve it.

    First, you can divide through by e^{2x} because it's never equal to 0.
    Then try to obtain \tan(x) one side and hence show the equality in question.
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    2cosx-sinx=0 ?
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    (Original post by poppydoodle)
    2cosx-sinx=0 ?
    Yes but as I said, rearrange to get tan(x)
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    (Original post by RDKGames)
    Yes but as I said, rearrange to get tan(x)
    I don’t understand how to do that
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    Hint:

     \displaystyle \tan(x) = \frac{\sin(x)}{\cos(x)}.
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    (Original post by poppydoodle)
    I don’t understand how to do that
    Well, look how tan(x) is defined in terms of sin(x) and cos(x), then think about what needs to happen to your equation if you want to obtain tan(x) in there.
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    Ahhh ok got it thank you
    For B do you set dy/dx to 0
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    (Original post by poppydoodle)
    Ahhh ok got it thank you
    For B do you set dy/dx to 0
    No because dy/dx=0 only for stationary points. Here is it asking you to find an equation of a tangent line through a point whose x-coordinate is x=0. So, you need to determine the gradient x=0, and also find the y-coordinate of this point before constructing the equation.
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    Could you explain it more? Not sure what you mean
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    (Original post by poppydoodle)
    Could you explain it more? Not sure what you mean
    You have a curve, there is a point with an x-coordinate equal to 0 on it, and you want to find the tangent line to the curve through it.

    In C1 you must've done problems where you find a tangent line to a curve through a point by finding the gradient of the curve at that point and constructing the equation of the tangent line. Same idea here.
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    So substitute 0 as x in dy/dx to find the gradient?
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    (Original post by poppydoodle)
    So substitute 0 as x in dy/dx to find the gradient?
    Yes.
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    So the gradient is 2?
    x=0, y= e^0 cos0
    x=0, y=1
    (0,1)
    y-1=2(x-0)

    y=2x+1

    Is this right?
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