Question Video: Finding the Derivative of a Polynomial Function Using the Definition of Derivative | Nagwa Question Video: Finding the Derivative of a Polynomial Function Using the Definition of Derivative | Nagwa

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

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Find the derivative of the function π(π₯) = 6π₯Β³ β 7π₯Β² using the definition of derivative, and then state the domain of the function and the domain of its derivative.

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### Video Transcript

Find the derivative of the function π of π₯ equals six π₯ cubed minus seven π₯ squared using the definition of derivative, and then state the domain of the function and the domain of its derivative.

There are two parts to the question, number one and number two. Letβs start by answering the first part of the question. This part of the question asks us to differentiate the function π. In other words, find a function which measures the steepness, or slope, of the curve given by π using a particular method, namely the definition of derivative. This means we should not use any other method, such as the standard formula for differentiating polynomials to arrive at the answer.

What is the definition of derivative? The definition of the derivative of a function π is the limit as β tends to zero of the expression π of π₯ plus β minus π of π₯ all divided by β. And itβs denoted by π dash of π₯, also known as π prime of π₯.

Letβs work out π of π₯ plus β. This is just π₯ replaced by π₯ plus β in π, which turns out to be six times π₯ plus β cubed minus seven times π₯ plus β squared. Letβs substitute this into the definition of the derivative of π. Letβs also substitute the expression for the function π in the definition of the derivative of π.

Now if we substitute β equals zero in an attempt to evaluate this limit, then we obtain the indeterminate form zero over zero, which is undefined. So, we cannot substitute β equals zero. In order to avoid the indeterminate form, letβs try to factorise out the term β in the numerator so that it will cancel out with the term β in the denominator.

Letβs start by factorising the expression for π of π₯ plus β. Firstly, note that π₯ plus β cubed is equal to π₯ plus β squared multiplied by π₯ plus β. Now note that the term π₯ plus β squared is common. So, we can factorise it out, obtaining π₯ plus β squared multiplied by six times π₯ plus β minus seven. We can expand out some of the parentheses to obtain π₯ squared plus two π₯β plus β squared all multiplied by six π₯ plus six β minus seven.

Now multiplying each term in the first parentheses by each term in the second parentheses, we obtain six π₯ cubed plus six π₯ squared β minus seven π₯ squared plus 12π₯ squared β plus 12π₯β squared minus 14π₯β plus six π₯β squared plus six β cubed minus seven β squared. Now we can collect together the like terms. Six π₯ squared β and 12π₯ squared β, as well as 12π₯β squared and six π₯β squared, to obtain six π₯ cubed plus 18π₯ squared β minus seven π₯ squared plus 18π₯β squared minus 14π₯β plus six β cubed minus seven β squared.

Letβs substitute this expansion of π of π₯ plus β into the definition of the derivative of π. Now letβs distribute the minus sign over the second parentheses in the limit. Doing so, we obtain the following limit. Now we can cancel out the terms six π₯ cubed and minus six π₯ cubed, as well as the terms minus seven π₯ squared and plus seven π₯ squared, to obtain the limit, as β tends to zero, of 18π₯ squared β plus 18π₯β squared minus 14π₯β plus six β cubed minus seven β squared all divided by β.

Noticed that the cancelling of the terms in the previous step has left us with a sum of terms in the numerator all of which contain the factor of β. This means we can factorise out the term β from the numerator as follows. We can now cancel out the term β in the numerator and denominator to obtain the following limit, in which we can safely substitute β equals zero.

Doing so, we obtain 18π₯ squared plus 18π₯ multiplied by zero minus 14π₯ plus six multiplied by zero squared minus seven multiplied by zero. This leaves us with two nonzero terms, 18π₯ squared minus 14π₯. So, using the definition of derivative, we have found that the derivative of the function π is 18π₯ squared minus 14π₯. This completes the answer to the first part of the question.

In order to answer the second part of the question, recall that the domain of a function of π₯ is the set of all π₯-values for which there exist π¦-values. For example, the domain of the function π¦ equals π₯ squared is the set of all real numbers. This is because for every real number π₯, the π¦-value π₯ squared exists.

Letβs have a look at another example. The domain of the function π¦ equals log π₯ is the set of all positive real numbers. This is because for every positive real number π₯, the π¦-value log π₯ exists. Note that if π₯ is less than or equal to zero, then log of π₯ is undefined.

Letβs sketch the function π. Factorising π, we obtain π of π₯ equals π₯ squared multiplied by six π₯ minus seven. So, π is a cubic function with a repeated root at π₯ equals zero and a single root at π₯ equals seven over six. So, it looks like this.

Recall that the directions of the right-hand side and left-hand side tails of a cubic function are determined by the sign of the cubic term. If the sign of the cubic term is positive, then the right-hand side tail would grow upwards since if π₯ is a large positive number, then so is π₯ cubed multiplied by a positive coefficient. On the other side, the left-hand side tail will grow downwards since if π₯ is a large negative number, then π₯ squared is a large positive number. And so, multiplying it by π₯, a large negative number, to obtain the cubic term, gives a large negative number, when the coefficient of the cubic term is positive.

We can do a similar analysis if the coefficient of the cubic term is negative. In the case of the cubic function π given to us in the question, the coefficient of the cubic term, which is six, is positive. So, the right-hand side tail grows upwards, and the left-hand side tail grows downwards. From the sketch, we can see that the domain of π is the set of all real numbers.

Letβs now sketch the derivative of π. The derivative of π is the quadratic function 18π₯ squared minus 14π₯ which factorises to two π₯ multiplied by nine π₯ minus seven. This quadratic function has two distinct roots, π₯ equals zero and π₯ equals seven over nine. So, its sketch is a parabola passing through zero and seven over nine. The parabola opens upwards as the coefficient of the π₯ squared term, which is 18, is positive. From the sketch, we can see that the domain of the derivative of π is also the set of all real numbers. This completes the answer to the second part of the question.

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