Find d^2y/dx^2 for the curve given by x = t^2 and y = t^2 + t.?

t = (x)^1/2

substitute t in y,

y = x + x^1/2

dy/dx = 1 + 1/(2x^1/2)

d^2y/dx^2 = -1/(4x^3/2) = -1/(4t^3)

Answer is (A)

y = x + sqrt(x)

dy/dx = 1 + 1/2(x^-1/2)

d2y/dx2 = -1/4 x^-3/2

= -1/4 t^-3

(a)

I don't want to just give you a letter, but i can give you some advice

if you know your basic calcules, you can do dy/dt and dx/dt easily.

Remember that dy/dx= dy/dt * dt/dx

of dy/dx= dy/dt / dx/dt

So you can get a function for dy/dx

And d^2y/dx^2 is just what you get if you take that function, and differentiate it w.r.t. x

x = t ²

dx / dt = 2t

d ² x / dt ² = 2

y = t ² + t

dy / dt = 2t + 1

d ² y / dt ² = 2

d ² y / dx ²

= (d ² y / dt ²) / (d ² x / dt ²) = 2 / 2 = 1

NONE of these is answer.

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a?