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    Can someone help me with these questions? this is from Edexcel C4 textbook ex.4B and 4c

    Find an expression in terms of x and y for dy/dx, given that square root of xy +x+y^2=0.

    Find dy/dx of y=xa^x and y=2^x/x

    It'd be helpful if you could put brackets round stuff mate. I know from Calculus at uni that when differentiating a function including a product xy, treat y as any other constant. Hope that helps

    There's something called 'implicit differentiation' where you don't start with a y = ...

    so you can differentiate xy with respect to x by treating as a product and remembering that you get dy/dx when you differentiate y.

    You can do your expression by using product and chain rules. Think of y^2 as a function-of-a-function and use chain rule

    (Original post by maltsheys)
    Find an expression in terms of x and y for dy/dx, given that square root of xy +x+y^2=0.
    \sqrt {xy + x + y^2} = 0

    xy + x + y^2 = 0

    x\frac {dy}{dx} + y + 1 + 2y\frac {dy}{dx} = 0

    \frac {dy}{dx}(x + 2y) = - y - 1

    \frac {dy}{dx} = \frac {-y - 1}{x + 2y}
    (Original post by maltsheys)
    Find dy/dx of y=xa^x
    lny = ln(xa^x) = lnx + ln(a^x) = lnx + xlna

    \frac {1}{y}\frac {dy}{dx} = \frac {1}{x} + lna

    \frac {dy}{dx} = y(\frac {1}{x} + lna)

    \frac {dy}{dx} = xa^x(\frac {1}{x} + lna)
    (Original post by maltsheys)
    xy = 2^x

    ln(xy) = ln(2^x)

    lnx + lny = xln2

    \frac {1}{x} + \frac {1}{y}\frac {dy}{dx} = ln2

    \frac {dy}{dx} = y(ln2 - \frac {1}{x})
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Updated: January 29, 2010
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