Question Video: Finding the Derivative of a Rational Function Using the Limit Definition of Derivatives | Nagwa Question Video: Finding the Derivative of a Rational Function Using the Limit Definition of Derivatives | Nagwa

# Question Video: Finding the Derivative of a Rational Function Using the Limit Definition of Derivatives Mathematics • Second Year of Secondary School

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Using the definition of a derivative, evaluate π/ππ₯ (1/(π₯ + 1)).

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### Video Transcript

Using the definition of a derivative, evaluate π by ππ₯ of one over π₯ plus one.

By definition, the derivative of π of π₯ β π by ππ₯ of π of π₯ β is the limit of π of π₯ plus β minus π of π₯ over β as β approaches zero. This is also written as π dash of π₯ or π prime of π₯ to emphasize that this is a function of π₯. Letβs apply this definition to our problem.

The denominator β we copy from the definition. Itβs also hopefully clear that π of π₯ should be one over π₯ plus one for our question. Certainly, it is on the left-hand side of the equation. Slightly more tricky perhaps is what π of π₯ plus β is. Well, if π of π₯ is one over π₯ plus one, then π of π₯ plus β is one over π₯ plus β plus one.

And now, weβre ready to find this limit. Itβs important to note that this is a limit as β approaches zero. This is not the limit as π₯ approaches some value. And our final answer will be in terms of π₯. If we try to direct substitution of β equals zero here, we will get the indeterminate form zero over zero. If you donβt believe me, you can check.

Weβre going to have to simplify the fraction in the limit and hopefully find a factor of β in the numerator to cancel with the β in the denominator before we can evaluate the limit. Currently, we have fractions in our fraction. There are two fractions in the numerator. If we multiply both numerator and denominator by π₯ plus β plus one, weβll get rid of one of those fractions.

Here, weβve distributed π₯ plus β plus one over the two terms in the numerator. And so we can recognize the first fraction as one. To get rid of the other fraction in the numerator, we multiply by π₯ plus one over π₯ plus one. And again, we simplify. So now our denominators become much more complicated, but our numerator is much simpler. And in fact, it can be simplified further.

We can get rid of the unnecessary parentheses around π₯ plus β for a start. And we notice that two π₯s cancel. We have an π₯, from which we then subtract an π₯. And something similar happens with the ones. So weβre just left with minus β in the numerator. And thereβs a factor of β in the denominator as well. Cancelling these factors, the numerator becomes negative one and the denominator is π₯ plus β plus one times π₯ plus one.

And having cancelled this factor of β in the numerator and the denominator, we can now directly substitute. Substituting zero for β, we get minus one over π₯ plus zero plus one times π₯ plus one. And of course, we donβt really need to write this plus zero explicitly. We get minus one over π₯ plus one times π₯ plus one therefore or minus one over π₯ plus one squared. So this is our answer.

Using the definition of a derivative, weβve shown that π by ππ₯ of one over π₯ plus one is minus one over π₯ plus one squared. As promised, this derivative is a function of π₯.

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