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Question Video: Differentiating a Function Involving Rational Functions at a Point Using the Quotient Rule Mathematics • Higher Education

Evaluate d𝑦/dπ‘₯ at (1, βˆ’1) if 𝑦 = βˆ’((2π‘₯)/(√(3π‘₯Β² + 1))).

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Video Transcript

Evaluate the first derivative of 𝑦 with respect to π‘₯ at the point one, negative one if 𝑦 is equal to negative two π‘₯ divided by the square root of three π‘₯ squared plus one.

In this question, we’re asked to evaluate d𝑦 by dπ‘₯ at a point and we’re given the function 𝑦. So to do this, we need to start by finding an expression for d𝑦 by dπ‘₯. And to do this, we start by noticing that 𝑦 is the quotient of two functions. So we’ll do this by using the quotient rule. We recall the quotient rule tells us the derivative of the quotient of two functions 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to 𝑓 prime of π‘₯ multiplied by 𝑔 of π‘₯ minus 𝑔 prime of π‘₯ multiplied by 𝑓 of π‘₯ all divided by 𝑔 of π‘₯ all squared. And this will be valid at all values of π‘₯ provided 𝑓 is differentiable at π‘₯, 𝑔 is differentiable at π‘₯, and 𝑔 of π‘₯ is not equal to zero. And since we’re asked to evaluate d𝑦 by dπ‘₯ at a point, we will need to check all of these conditions hold.

So we’ll start by setting 𝑓 of π‘₯ to be the function in our numerator, negative two π‘₯, and 𝑔 of π‘₯ to be the function in the denominator, the square root of three π‘₯ squared. And we notice we’re asked to evaluate d𝑦 by dπ‘₯ at the point one, negative one. That’s when π‘₯ is equal to one. So to validate that we can use the quotient rule, we need to check that 𝑓 is differentiable at one, which we know it is because 𝑓 is a linear function. It’s differentiable at all values of π‘₯. Second, we need to check that 𝑔 is differentiable at one. But before we do this, let’s move on to the third condition. We need 𝑔 of one to not be equal to zero.

We can check this by substituting π‘₯ is equal to one into our function 𝑔 of π‘₯. We get 𝑔 of one is the square root of three times one squared plus one, which we can calculate is equal to two. So 𝑔 of one is not equal to zero, and the third condition for the quotient rule is true. All we need now is 𝑔 to be differentiable at π‘₯ is equal to one. And there are a few different ways of doing this. The easiest way is just to find an expression for 𝑔 prime of π‘₯. To do this, we notice that 𝑔 is the composition of two functions. So we’ll do this by using the chain rule.

We recall the chain rule tells us the derivative of the composition of two functions 𝑒 evaluated at 𝑣 of π‘₯ is equal to 𝑣 prime of π‘₯ multiplied by 𝑒 prime evaluated at 𝑣 of π‘₯. And this is true for all values of π‘₯ provided 𝑣 is differentiable at π‘₯ and 𝑒 is differentiable at 𝑣 evaluated at π‘₯. We’ll set our function 𝑣 to be the inner function. That’s three π‘₯ squared plus one. We can then set 𝑒 to be our outer function. 𝑒 evaluated at 𝑣 is the square root of 𝑣. Then we see that 𝑔 of π‘₯ is equal to 𝑒 evaluated at 𝑣 of π‘₯. 𝑒 is a function of 𝑣, and 𝑣 in turn is a function in π‘₯. So we can differentiate this by using the chain rule.

To do this, let’s start by finding an expression for 𝑣 prime of π‘₯. That’s the derivative of a polynomial. So we’ll do this term by term by using the power rule for differentiation. We multiply by the exponent of π‘₯ and reduce this exponent by one. 𝑣 prime of π‘₯ is six π‘₯. We can do exactly the same to find an expression for 𝑒 prime of 𝑣. We start by rewriting the square root of 𝑣 as 𝑣 to the power of one-half, and then we apply the power rule for differentiation. We multiply by the exponent of 𝑣 and reduce this exponent by one. 𝑒 prime of 𝑣 is one-half times 𝑣 to the power of negative one-half, which we can rewrite as one divided by two root 𝑣.

But remember, in our question, we’re evaluating d𝑦 by dπ‘₯ at π‘₯ is equal to one. So we need to check that 𝑣 is differentiable when π‘₯ is equal to one. And we need to check that 𝑒 is differentiable at 𝑣 evaluated at one. Let’s start by checking that 𝑣 is differentiable when π‘₯ is equal to one. 𝑣 of π‘₯ is a polynomial, and polynomials are differentiable for all real numbers. So 𝑣 is differentiable when π‘₯ is equal to one. Next, let’s check that 𝑒 is differentiable at 𝑣 evaluated at one. We’ll start by finding 𝑣 evaluated at one by substituting π‘₯ is equal to one into 𝑣 of π‘₯. 𝑣 of one is three times one squared plus one, which is equal to four.

So we need to check that our function 𝑒 evaluated at 𝑣 is differentiable when 𝑣 is equal to four, and 𝑒 of 𝑣 is the square root function. This is differentiable for all positive values of 𝑣. So it is differentiable at 𝑣 evaluated at one. Therefore, the chain rule will be valid when π‘₯ is equal to one, and we can use the chain rule to find 𝑔 prime of π‘₯. 𝑔 is differentiable when π‘₯ is equal to one. We just need to substitute our expressions for 𝑣 prime, 𝑒 prime, and 𝑣 into the chain rule. Doing this, we get that 𝑔 prime of π‘₯ is equal to six π‘₯ multiplied by one divided by two root three π‘₯ squared plus one. And we can simplify this expression. We can cancel the shared factor of two in the numerator and denominator. This then gives us that 𝑔 prime of π‘₯ is equal to three π‘₯ divided by the square root of three π‘₯ squared plus one.

And before we apply the quotient rule, it’s worth pointing out when we tried to apply the chain rule, we showed that 𝑔 was differentiable when π‘₯ is equal to one. This was the final condition we needed to confirm that we could apply the quotient rule when π‘₯ was equal to one. Therefore, we can use the quotient rule to find an expression for d𝑦 by dπ‘₯. And this expression will be valid when π‘₯ is equal to one. To apply the quotient rule, we need to find expressions for 𝑓, 𝑔, 𝑔 prime, and 𝑓 prime. We have expressions for all of these except 𝑓 prime of π‘₯. We can find an expression for 𝑓 prime of π‘₯ by differentiating 𝑓 of π‘₯ with respect to π‘₯. 𝑓 is a linear function, so its derivative is the coefficient of π‘₯, which is negative two.

Now we could substitute our expressions for 𝑓 prime, 𝑔, 𝑔 prime, and 𝑓 into the quotient rule. However, this would construct a very complicated equation. Instead, remember, we only need to evaluate this at π‘₯ is equal to one. So instead, let’s evaluate the quotient rule at π‘₯ is equal to one. We have d𝑦 by dπ‘₯ at π‘₯ is equal to one is equal to 𝑓 prime of one times 𝑔 of one minus 𝑔 prime of one times 𝑓 of one all divided by 𝑔 of one all squared. And it’s much easier to just evaluate these individual functions at one and then substitute these values into the expression. For example, we can find 𝑓 prime of one by substituting π‘₯ is equal to one into the function 𝑓 prime of π‘₯. But 𝑓 prime is a constant function, so 𝑓 prime of one is just equal to negative two. And we can keep a note of this in the top-right corner of our page.

Next, we’ll find the value of 𝑔 of one by substituting π‘₯ is equal to one into our function 𝑔 of π‘₯. We actually already found this value when we were checking that the quotient rule was valid. 𝑔 evaluated at one is the square root of three times one squared plus one, which is the square root of four, which is equal to two. Once again, we’ll keep a note of this in the top-right corner of our page, and then we’ll find 𝑔 prime of one. We substitute π‘₯ is equal to one into our function 𝑔 prime of π‘₯. 𝑔 prime of one is three times one divided by the square root of three multiplied by one squared plus one. And if we evaluate this expression, we get 𝑔 prime of one is three over two. We can then keep a note of this. And finally, the only value left is 𝑓 evaluated at one.

We substitute π‘₯ is equal to one into our function 𝑓 of π‘₯. We get 𝑓 of one is negative two times one, which is equal to negative two. Now, all we need to do is substitute these four values into our quotient rule formula evaluated at π‘₯ is equal to one. We get d𝑦 by dπ‘₯ at π‘₯ is equal to one is equal to negative two times two minus three over two multiplied by negative two all divided by two squared. We can then simplify this expression. First, we have negative two times two is equal to negative four. Next, negative three over two multiplied by negative two is equal to positive three. Finally, two squared is equal to four. So this simplifies to give us negative four plus three all divided by four, which we can evaluate is equal to negative one-quarter, which is our final answer.

Therefore, we were able to show if 𝑦 is equal to negative two π‘₯ divided by the square root of three π‘₯ squared plus one, then d𝑦 by dπ‘₯ evaluated at the point one, negative one is equal to negative one-quarter.

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