# Question Video: Differentiating a Quotient of Functions Involving Root and Polynomial Functions

Find the first derivative of π¦ = (β3π₯Β² β 2π₯ + 17)/β(π₯) with respect to π₯.

02:14

### Video Transcript

Find the first derivative of π¦ equals negative three π₯ squared minus two π₯ plus 17 over the square root of π₯ with respect to π₯.

Letβs look carefully at what weβre trying to differentiate. We can see itβs the quotient of two differentiable functions. And so weβre going to apply the quotient rule. This says that the derivative of π’ divided by π£ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. So our first job is to define π’ and π£.

Since π’ is the numerator of our fraction, we define it to be equal to negative three π₯ squared minus two π₯ plus 17. And weβll let π£ be equal to root π₯. Now in fact, we know that the square root of π₯ can be written as π₯ to the power of one-half. This will make the next step easier.

We need to find dπ’ by dπ₯ and dπ£ by dπ₯. And so we recall that, to differentiate a polynomial term, we multiply the entire term by the exponent and then reduce that exponent by one. This means the derivative of negative three π₯ squared is two times negative three π₯, which is negative six π₯. Then the derivative of negative two π₯ is negative two. And we know that when we differentiate a constant, we get zero. When we differentiate π₯ to the power of one-half, we get a half times π₯ to the power of negative one-half.

And weβre ready to substitute everything we know into our formula for the quotient rule. Itβs π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. Letβs simplify this fraction. Weβll first multiply both the numerator and the denominator by two π₯ to the power of one-half. Then our fraction becomes two π₯ multiplied by negative six π₯ minus two minus negative three π₯ squared minus two π₯ plus 17. All over two π₯ to the power of one-half times π₯ to the power of one-half squared.

Letβs rewrite our denominator as two π₯ to the power of one-half times π₯ to the power of one-half times π₯ to the power of one-half. And then we recall that we can simply add the exponents. So we get two π₯ to the power of three over two, which can alternatively be written as two π₯ cubed to the power of one-half. Which is two times the square root of π₯ cubed. By distributing the parentheses on our numerator, we get negative two π₯ squared minus four π₯ plus three π₯ squared plus two π₯ minus 17.

And finally, we simplify. And we see that the first derivative of our function with respect to π₯ is negative nine π₯ squared minus two π₯ minus 17 over two times the square root of π₯ cubed.