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Math Help - calculus with exponents

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    calculus with exponents

    I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
    On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.

    a) Show that the first derivative=e^(2x)(2cosx-sinx)
    b) Find second derivative

    At the point P(a,b), the second derivative = 0.
    c) Use results from a) and b) to prove that
    i) tan a = 3/4
    ii) the gradient of the curve at P is e^(2a)

    Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).

    If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
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  2. #2
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    Quote Originally Posted by bhuang View Post
    I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
    On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.

    a) Show that the first derivative=e^(2x)(2cosx-sinx)
    b) Find second derivative

    At the point P(a,b), the second derivative = 0.
    c) Use results from a) and b) to prove that
    i) tan a = 3/4
    ii) the gradient of the curve at P is e^(2a)

    Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).

    If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
    you know \tan{a} = \frac{3}{4} ... so, what would be the values of \sin{a} and \cos{a} ? ... and how will those values determine the value of e^{2a}(2\cos{a}-\sin{a}) ?
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