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

Find the first derivative of 𝑦 = (βˆ’3π‘₯Β² βˆ’ 2π‘₯ + 17)/√(π‘₯) with respect to π‘₯.

02:14

Video Transcript

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

Let’s look carefully at what we’re trying to differentiate. We can see it’s the quotient of two differentiable functions. And so we’re going to apply the quotient rule. This says that the derivative of 𝑒 divided by 𝑣 is equal to 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. So our first job is to define 𝑒 and 𝑣.

Since 𝑒 is the numerator of our fraction, we define it to be equal to negative three π‘₯ squared minus two π‘₯ plus 17. And we’ll let 𝑣 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 d𝑒 by dπ‘₯ and d𝑣 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 we’re ready to substitute everything we know into our formula for the quotient rule. It’s 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. Let’s simplify this fraction. We’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.

Let’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.

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