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    I'm stuck on the following questions:

    1. y= 5x^2 - 10x + 9 / (x-1)^2

    Show that dy/dx = 8 / (x-1)^2

    I used the product rule to give me:

    (x-1)².(10x-10) - (5x^2-10x+9).(2(x-1)) / (x-1)^4

    But i can't simplify this further to give me the answer

    2. Given that x=Tan 1/2y prove that dy/dx= 2/1+x²

    I tried several things but i couldn't eliminate the Tan

    3. f(x) = x + (3/x-1) - (12/x²+2x-3)

    Show that f(x) = x²+3x+3 / x+3

    I've tried getting a common denominator but i ended up with the wrong answer!

    Can anyone help me please?

    1. y= 5x^2 - 10x + 9 / (x-1)^2

    Show that dy/dx = 8 / (x-1)^2
    This one - do you mean

     y= 5x^2 - 10x + \frac{9}{(x-1)^2}

    If so there should be no reason to use product rule?

    2. Given that x=Tan 1/2y prove that dy/dx= 2/1+x²
    This one - try finding  \frac{d\mathbf{x}}{d\mathbf{y}} and inverting this to get  \frac{dy}{dx}

    Also the identity

     sec^2(t) = 1 + tan^2(t)

    might come in handy...

    3. f(x) = x + (3/x-1) - (12/x²+2x-3)
    Useful thing to notice here is that

     x^2+2x-3 = (x-1)(x+3)

    That might help...

    Hey Ridium, ok lets see if I can be of some help here ;D

    1. Ok for this question, notice that you have a quadratic over a quadratic

    \displaystyle \frac{5x^2 - 10x + 9}{x^2 - 2x + 1}

    Now its would be simpler to simlify this first, this is a 'top heavy' polynomial fraction, how can you simplify this, familiar with polynomial long division, I belive you learn that in C3 . Also I dont think that the end result is correct, I think it should be:

    \displaystyle \frac{dy}{dx} = -\frac{8}{(x-1)^3}

    does it actually say show the equation you quoted?

    2. Ok with this one you actually no not have to eliminate tan from the equations, this is implicit differentiation, a term you hopefully have come accross before, consider this:

    \displaystyle\frac{d}{dx}(x) = \frac{d}{dx}\left(tan\left( \frac{y}{2}\right)\right) = \frac{d}{dy}\left(tan\left( \frac{y}{2}\right)\right)\frac{d  y}{dx}

    can you see how I have applied the chain rule here, imagine rite now that your not differentiaing one side of an equation but individual terms, whats the diriviate of x with respect to x, whats the derivative of tan 1/2y wrt y. have a think. (EDIT: spread_logic's why is much simpler than mine for a method to question 2 if you dont understand this way )

    3. Ok first can you factorize x^2 + 2x -3, could (x-1) be a factor, ;D I cant say anymore on that.

    Have a think about that, then have a go at trying to solve them again
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Updated: February 7, 2010

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