Question

What is (dy/dx)^2=dy^2/dx^2

Answer 1

it's wrong according to differential caculus .
(dy/dx)² ≠ d²y/dx²

Let us see one example :-
y = sinx
now differentiate with respect to x
dy/dx = cosx
take square both sides,
(dy/dx)² = cos²x

dy/dx = cosx
again, differentiate with respect to x
d{dy/dx}/dx = -sinx
d²y/dx² = - sinx

you can see that ,
(dy/dx)² ≠ d²y/dx²

actually , (dy/dx)² means square of first derrivative of function y . but d²y/dx² means 2nd derrivatives of given function.
so, it can't true statement ,
(dy/dx)² ≠ d²y/dx²

Answer 2

Step-by-step explanation:

The second derivative can be used as an easier way of determining the nature of stationary points (whether they are maximum points, minimum points or points of inflection).

A stationary point on a curve occurs when dy/dx = 0. Once you have established where there is a stationary point, the type of stationary point (maximum, minimum or point of inflexion) can be determined using the second derivative.

If d2y is positive, then it is a minimum point

dx2

If d2y is negative, then it is a maximum point

dx2

If d2y = zero, then it could be a maximum, minimum or point of inflexion

dx2

If d2y/dx2 = 0, you must test the values of dy/dx either side of the stationary point, as before in the stationary points section.

Example

Find the stationary points on the curve y = x3 - 27x and determine the nature of the points:

At stationary points, dy/dx = 0

dy/dx = 3x2 - 27

If this is equal to zero, 3x2 - 27 = 0

Hence x2 - 9 = 0 (dividing by 3)

So (x + 3)(x - 3) = 0

So x = 3 or -3

d2y/dx2 = 6x

When x = 3, d2y/dx2 = 18, which is positive.

When x = -3, d2y/dx2 = -18, which is negative.

Hence there is a minimum point at x = 3 and a maximum point at x = -3.