Question

How do you find the derivative of `f(x)=x^2-5x+3`?

Answer 1

`f'(x)=2x-5`

Explanation:

The derivative of a function is the instantaneous rate of change of the function at a given point. But we have a function, so we want to find the rate of change throughout. We have two ways of doing this, and I'll show you both ways.

Method 1: The Power Rule

This method only works for polynomials (`x^n`). We have a saying that helps remember the power rule: "Bring the power forward, take one off the power." Essentially, that means the following:

`d/dx(x^n)=nx^(n-1)`

So we can apply this to your function. But we also have a constant. Remember that adding or subtracting a constant yields only a vertical shift, not a stretch. Therefore, it has no effect on the rate of change, so we can also say:

`d/dx(n)=0`

We can use these two rules together to get the answer:

`d/dx(x^2-5x+3)=2x-5`

Method 2: First Principles

First Principles is a method that allows you to take the derivative of any function. It comes from the slope formula (`(rise)/(run)`). We want to get to an infinitely small interval on the graph, so we can use the slope formula to come up with the slope at a given point of the function:

`lim_(h->0)((f(x+h)-f(x))/h)`

So let's plug in your function:

`lim_(h->0)((((x+h)^2-5(x+h)+3)-(x^2-5x+3))/h)`

Since we can't divide by 0, we need to get the h out of the denominator. So let's expand the binomial.

`lim_(h->0)(((x^2+2xh+2h^2-5x-5h+3)-(x^2-5x+3))/h)`

And let's combine all the like terms (cancelling out most of the x values):

`lim_(h->0)((2xh+2h^2-5h))/h`

Now, every term as an h somewhere in it. So we can divide by h:

`lim_(h->0)(2x+2h-5)`

And now as the limit approaches 0, 2h reaches 0, and therefore, we are left with the final answer:

`f'(x)=2x-5`