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

Question

Scarlett Allison · Accepted Answer

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

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#

Charles Arocha · Answer

undefined To find the derivative of $ f(x) = x^2 - 5x + 3 $, you apply the power rule and constant rule. The derivative is $ f'(x) = 2x - 5 $.