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anonymous

  • 5 years ago

if x^1/2 +y^1/2=2,find d^2y/dx^2

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  1. anonymous
    • 5 years ago
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    ok this is implicit differentiation. you want to take the derivative with respect to x. d/(dx) of x^(1/2) is 1/(2(x^(1/2)) one over the quantity two square root of x. d/(dx) of y^(1/2) is 1/(2(x^(1/2)) dy/dx d/(dx) of 2 is a constant so it goes to 0. So your first equation for dy/dx = 1/(2(x^(1/2)) + 1/(2(y^(1/2)) dy/dx = 0 I'll do the second derivative d^2y/dx^2 in the next post

  2. anonymous
    • 5 years ago
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    1/(2(y^(1/2)) dy/dx = -(1/(2(x^(1/2)) ) dy/dx = -(1/(2(x^(1/2)) ) / (1/(2(y^(1/2))) take the derivative of that again^

  3. anonymous
    • 5 years ago
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    I have to start writing this down, hold on.

  4. anonymous
    • 5 years ago
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    dy/dx = -y^(1/2) / x^(1/2) Next use the quotient rule. (VdU-UdV)/V^2 -x^(1/2)/(2(y^1/2)) * dy/dx + (minus a negative) y^(1/2)/(2(x^(1/2)) all divided by x <-- (x^(1/2))^2 Now plug in dy/dx and that's your answer. Interestingly, both y^1/2 and x^1/2 in the left term of the numerator cancel with the first derivative's terms, as well as 2/2, leaving d^2y/dx^2 = (1/2 + y^(1/2) / (2*x^(1/2))) / x

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