Finding 1st and 2nd derivatives

I'm confused, I can't figure out how to find derivatives.
I am asked to find the 1st & 2nd derivative of Y=3x^2-12x-5x^-3
Could someone show me how to do this step by step please? The formulas are dy/dx for 1st derivative and d^2y/dx^2

Thanks

The easiest way to find a derivative is by using the power rule. If f(x) = a^n, then f'(x) = n*a^(n-1). In layman's terms, that means you take each term, multiply the power by the base and subtract the power by one. If the term is a constant, the derivative is zero.

So, to solve this problem,

Y = 3x^2 - 12x - 5x^-3

Y' (First Derivative) = 6x -12 - 15x^-4

Y'' (Second Derivative) = 6 - 0 - 60x^-5 = 6 - 60x^-5


Hope that helped!

i do not know where you are in your calculus class. if you have learned differentiation rules, then the simplest way to go about solving this problem is to let f(x)=3x^2-12x-5x^-3 and then use the difference rule, the constant multiple rule, and the power rule to solve.

Difference Rule: d/dx[f(x)-g(x)]=d/dx f(x)- d/dx g(x)
Constant Multiple Rule: d/dx[cf(x)]=cd/dxf(x)
Power Rule: d/dx(x^n)=nx^(n-1)

This gives you f '(x) =3(2)x^(2-1)-12-5(-3)x^(-3-1)==>
6x-12+15x^-4

If you have not yet learned differentiation rules then you need to use the definition of the derivative.

simply put, the derivative is the graph of the various SLOPES of a curve at the points along the curve. The slope of a curve is the rise over the run (you may recognize this as y2-y1/x2-x1, so the slope of a curve at a specific point is the rise/run as the distance between the two points approaches 0. so the derivative of an equation as a function is given by:

f '(x)=lim(h->0) {[f(x+h)-f(x)]/h}

as you can see, that is the rise (y2-y1=f(x-h)-f(x)) over the run (x2-x1=(x+h)-x=h).

lim(h->0) [3(x+h)^2-12(x+h)-5/(x+h)^3-(3x^2-12x-5x…