Maths Derivatives & Calulus

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1) Find d^2y / dx^2 given that y^3 - x^2 = 4


2)If y = xª + ( 1 / xª ) , prove that

x^2 . y'' + xy' – a^2 . y = 0


3) Find dy/dx for the following function.
cos ( x - y ) = ( y ) ( 2x + 1 ) ^3

4)Find the values of A and B that make f(x) differentiable at
x = 0

f(x) = A + 2x + 3, if x > 0
AND
f(x) = 5 + B (sin²x / |x|) , if x < 0

  • Maths Derivatives & Calulus -

    y^3 - x^2 = 4
    3y^2 y' - 2x = 0
    y' = 2x/3y^2

    6y y'^2 + 3y^2 y'' - 2 = 0

    y'' = (2 - 6yy'^2)/3
    = (2 - 6y(2x/3y^2)^2)/3
    _______________

    y = x^a + 1/x^a
    y' = ax^(a-1) - a/x^(a+1)
    y'' = a(a-1)x^(a-2) + a(a+1)/x^(a+2)

    x^2 y'' = a(a-1)x^a + a(a+1)/x^a
    xy' = ax^a - a/x^a
    -a^2 y = -a^2 x^a - a^2/x^a

    Add 'em up: 0
    _______________

    cos(x-y) = y(2x+1)^3
    -sin(x-y)(1-y') = y'(2x+1)^3 + 6y(2x+1)^2
    y'(sin(x-y) - y'(2x+1)^3 = xsin(x-y) + 6y(2x+1)^2
    carry on
    __________________

    lim sin^2 x/x = 2sincos/1 = 0

    f(0) = A + 3
    f(0) = 5 + 0B = 5

    So, if A=2, then the limit exists from both sides

  • Maths Derivatives & Calulus -

    thank u so much for ur effort and time
    helped me alot!!!!!

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