Maths Derivatives & Calulus
posted by Yousef .
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

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^(a1)  a/x^(a+1)
y'' = a(a1)x^(a2) + a(a+1)/x^(a+2)
x^2 y'' = a(a1)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(xy) = y(2x+1)^3
sin(xy)(1y') = y'(2x+1)^3 + 6y(2x+1)^2
y'(sin(xy)  y'(2x+1)^3 = xsin(xy) + 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 
thank u so much for ur effort and time
helped me alot!!!!!