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A more complete question is required to receive a good answer here. If you were to ask how to derive a function, I could take you through a general rundown of the basic derivative patterns of most common functions. Since you said "equation" though, this implies the possibility of implicit differentiation methods being required. If you were to narrow down your question more, you might get a better answer.

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Q: Find the derivative of an equation?
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How do you find the Min value for y in this equation?

You didn't specify the equation. A minimum or maximum value of a function is often found by calculating the derivative of a function, writing an equation for derivative equal to zero, and then analyzing points where the derivative either doesn't exist, or is equal to zero. You'll find find information about this in introductory calculus books.


If the 2nd derivative of an equation isn't constant is it still a quadratic relation?

No. A quadratic equation always has a second derivative that is a constant. For example -3x2 + 10x - 2 first derivative -6x + 10 second derivative -6


How the slope of a curved line at a point can be found?

The slope of a curved line at a point is the slope of the tangent to the curve at that point. If you know the equation of the curve and the curve is well behaved, you can find the derivative of the equation of the curve. The value of the derivative, at the point in question, is the slope of the curved line at that point.


How do you find the maximum or minimum of a parabola?

Solve for when the first derivative is equal to zero. If you don't know how to take a derivative, then put the equation into the form y = Ax2 + Bx + C. The derivative of this will be 2Ax + B, so at x = -B / (2*A), and y = -B2/(4*A) + C


Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?

pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!