Question Video: Find the Derivative of the Inverse of a Rational Function at a Point | Nagwa Question Video: Find the Derivative of the Inverse of a Rational Function at a Point | Nagwa

Question Video: Find the Derivative of the Inverse of a Rational Function at a Point Mathematics • Higher Education

For the function 𝑓(π‘₯) = π‘₯ βˆ’ (3/π‘₯) at π‘Ž = 2, π‘₯ > 0, find (𝑓⁻¹)β€²(π‘Ž).

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Video Transcript

For the function 𝑓 of π‘₯ is equal to π‘₯ minus three divided by π‘₯ at π‘Ž is equal to two and π‘₯ is greater than zero, find 𝑓 inverse prime evaluated at π‘Ž.

The question gives us a function 𝑓 of π‘₯, and it wants us to find the first derivative of the inverse function of 𝑓 evaluated at π‘₯ is equal to π‘Ž. And we know how to find the derivative of an inverse function at a point π‘₯ is equal to π‘Ž. The derivative of 𝑓 inverse at a point π‘₯ is equal to π‘Ž is equal to one divided by 𝑓 prime evaluated at 𝑓 inverse of π‘Ž. And this is provided π‘Ž is in the domain of our inverse function. And 𝑓 prime is not equal to zero at this point.

The question wants us to determine 𝑓 inverse prime at π‘₯ is equal to two. So we’ll substitute π‘Ž is equal to two into this expression. We see there’s two things we need to find to use this expression. We need to find an expression for 𝑓 prime of π‘₯, and we need to evaluate 𝑓 inverse of two. Let’s start by finding an expression for 𝑓 prime of π‘₯. First, we’ll rewrite 𝑓 of π‘₯ by using our laws of exponents. It’s equal to π‘₯ minus three times π‘₯ to the power of negative one. We can then find 𝑓 prime of π‘₯ by using the power rule for differentiation. We get one plus three times π‘₯ to the power of negative two. We now need to find an expression for our inverse function of 𝑓 evaluated at two. Let’s say that this is equal to some value 𝑏.

And we can use the definition of an inverse function to find an expression for 𝑏. First, we apply 𝑓 to both sides of this equation. And notice, since the domain of our function 𝑓 demands that π‘₯ is greater than zero, this tells us that 𝑏 must be greater than zero. Now, by using the definition of an inverse function, 𝑓 evaluated at 𝑓 inverse of two is just equal to two since a function composed with its inverse function should be the identity function. It won’t change our input. So we get two is equal to 𝑓 evaluated at 𝑏. And we know how to evaluate 𝑏. We’re given an expression for 𝑓 in the question. Substituting in π‘₯ is equal to 𝑏 into our expression for 𝑓 of π‘₯, we have 𝑓 evaluated at 𝑏 is equal to 𝑏 minus three over 𝑏.

So we have two is equal to 𝑏 minus three over 𝑏. And we know that 𝑏 is greater than zero. So we’ll just solve this equation to find our value of 𝑏. We’ll start by multiplying through by 𝑏. This gives us two 𝑏 is equal to 𝑏 squared minus three. Next, we’ll subtract two 𝑏 from both sides of this equation. This gives us that zero is equal to 𝑏 squared minus two 𝑏 minus three. This is a quadratic equation in terms of 𝑏. And there’s a few different ways of solving quadratic equations. For example, we could do this by using the quadratic formula. However, we could also notice that negative three multiplied by one is equal to negative three. And negative three plus one is equal to negative two.

So we can factor this quadratic to give us 𝑏 minus three times 𝑏 plus one. And we have to be careful at this point. Since we know 𝑏 must be greater than zero, 𝑏 cannot be equal to negative one. So our only solution to this quadratic is 𝑏 is equal to three. So our value of 𝑏 is three. In other words, the inverse function of 𝑓 evaluated at two is equal to three. So we’re now ready to find 𝑓 inverse prime evaluated at two. It’s equal to one divided by 𝑓 prime evaluated at the inverse function of 𝑓 evaluated at two. We already showed the inverse function of 𝑓 evaluated at two is equal to three. So our derivative just becomes one divided by 𝑓 prime evaluated at three. And we already found an expression for 𝑓 prime of π‘₯. It’s equal to one plus three times π‘₯ to the power of negative two.

So we’ll just substitute π‘₯ is equal to three into our expression for 𝑓 prime of π‘₯. We get one divided by one plus three times three to the power of negative two. We can simplify this to get one divided by one plus one-third, which then simplifies to give us one divided by four over three. And we take the reciprocal of this; we get three divided by four. Therefore, we’ve shown for the function 𝑓 of π‘₯ is equal to π‘₯ minus three over π‘₯, where π‘Ž is equal to two and π‘₯ must be greater than zero, then the derivative of the inverse function at π‘₯ is equal to π‘Ž must be equal to three divided by four.

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