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    • Thread Starter

    Hi All

    Please help a returning mature student...
    Help needed with 2 Differentiation Questions.

    Q= 2e^x + e^-2x
    A= 2e^x + -2e^-2x

    Why would they keep 2e^x in the answer ?
    I do understand the -2e^-2x part though managed to work that one out correctly..

    Q= sin(-8Q) the Q is supposed to be theta dont know how to write it on keyboard.
    A= I would tend to give -8 Cos -8Q but the actual answer is
    -8Cos(-8Q)=-8Cos 8Q. So I dont understand how they come to that please help.

    Kind regards

    When writing down answers, it's best not to use Q=... and A=... as this is not a standard notation so it can be confusing. Instead, you can use the notation \frac{d}{dx}f(x)=g(x) which means that if you differentiate f(x) you get g(x). So for example, you might write your answer to the first question as \dfrac{d}{dx}(2e^x+e^{-2x})=2e^x-2e^{-2x}.

    Anyway, to answer your question, when you differentiate 2e^x you can use the chain rule if you're not sure. So let y=2e^x and u=e^x so y=2u. Now \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}= 2 \times  e^x = 2e^x. So if we differentiate 2e^x we get 2e^x.

    For the second part, sketch a graph of y=cos(x). You'll notice that the graph is symmetrical about the y axis, so we could swap round the x and -x axes and still get the same graph. What this means mathematically is that cos(x)=cos(-x) which is a general trigonometric identity that is useful to know. So in your case, you have -8\cos(-8\theta) so using this identity we have, we can say that this is equal to -8\cos(8\theta). What this does is it eliminates an extra minus sign, making the answer slightly simpler.
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