Video: Differentiating Polynomial Functions

Differentiate the function 𝑓(π‘₯) = βˆ’π‘₯Β²/7 βˆ’ 3π‘₯ βˆ’ 4.

06:02

Video Transcript

Differentiate the function 𝑓 of π‘₯ equals minus π‘₯ squared over seven minus three π‘₯ minus four.

When we differentiate 𝑓 of π‘₯, we get another function, the function 𝑓 prime of π‘₯. The function 𝑓 prime gives the slope of the tangent to the graph of 𝑓 for a given value of π‘₯. Okay, so let’s differentiate the expression we have for 𝑓 of π‘₯. One property of differentiation that we can use to find this is that the derivative of a sum or difference of two functions is the sum or difference, as appropriate, of the derivatives of the two functions. And in fact, this isn’t just true for the sum or difference of two functions. This is true for the sum or difference of any finite number of functions. As a result, we can consider the derivatives of minus π‘₯ squared over seven, three π‘₯, and four separately.

Let’s start by finding the derivatives of the function that takes π‘₯ to minus π‘₯ squared over seven. The derivative of this function is another function which outputs the slope of the tangent to the function minus π‘₯ squared over seven. Here is a quick sketch of the graph of the function we want to differentiate. It’s a downward-facing parabola with vertex at the origin. Now, what can we say about its derivative? Well, for this value of π‘₯, the tangent to the graph is as shown. We see that it has positive slope, perhaps with a value of about one. So for this value of π‘₯, the derivative of our function 𝑔 of π‘₯ is one. When π‘₯ is zero, the tangent lies on the π‘₯-axis which has a slope of zero. And so, the derivative of the function 𝑔 of π‘₯ is zero for this value of π‘₯.

Unfortunately, we can’t just read off the derivative 𝑑 by 𝑑π‘₯ of minus π‘₯ squared over seven from this graph. Instead, we use a couple of properties or rules of differentiation, that the derivative of a number times a function is that number times the derivative of the function. And so, the derivative of negative seventh π‘₯ squared is negative seventh times the derivative of π‘₯ squared. And we also use the fact that the derivative of π‘₯ to the power of 𝑛 with respect to π‘₯ is 𝑛 times π‘₯ to the power of 𝑛 minus one. And so, 𝑑 by 𝑑π‘₯ of π‘₯ to the power of two is two times π‘₯ to the power of one or just two π‘₯.

Putting these two together, we get negative a seventh times two π‘₯ which is negative two sevenths π‘₯. You might like to pause the video now. Have a look at the sketch and convince yourself that a derivative of negative two sevenths π‘₯ is reasonable.

Okay, what about the other terms? What is 𝑑 by 𝑑π‘₯ of three π‘₯? Can you see what the slope is here? We could apply the rule about the derivative of a number times a function being that number times the derivative of the function. We could also apply the fact that the derivative of π‘₯ with respect to π‘₯ is just one. You might remember that directly or see it as an application of the power rule with 𝑛 equal to one. In any case, we get an answer of three. Remember that we have to subtract this from the previous term. Now again, you can look at the graph and ask yourself if it makes sense for the derivative of three π‘₯ to be three.

Finally, we need to find the derivative of the constant function four. Here, we can use the fact that the derivative of any constant function π‘˜, which is just some number, is zero. And so, 𝑑 by 𝑑π‘₯ of four is zero. Again, you can think about why this makes sense, given the graph of the constant function four.

Tidying up then, we get our final answer that the derivative 𝑓 prime of π‘₯ is negative two sevenths π‘₯ minus three. Here, we used various differentiation rules to find this derivative. But for such a simple function, we can actually do this from first principles.

By definition, the derivative 𝑓 prime of π‘₯ is the limit of 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ all over β„Ž, as β„Ž approaches zero. Now, what is 𝑓 of π‘₯ plus β„Ž? Well, 𝑓 of π‘₯ is minus π‘₯ squared over seven mins three π‘₯ minus four. So 𝑓 of π‘₯ plus β„Ž is minus π‘₯ plus β„Ž squared over seven minus three times π‘₯ plus β„Ž minus four. Substituting the expression for 𝑓 of π‘₯ completes the numerator. And we just have β„Ž in the denominator.

We can expand the parentheses in the numerator. And we can then note that lots of cancellation occurs. The remaining terms in the numerator have a common factor of β„Ž. Factoring β„Ž from these terms, we see that we can cancel the β„Ž in the denominator. We’re left with a limit that we can evaluate by directly substituting zero for β„Ž. Doing this, we find that the derivative 𝑓 prime of π‘₯ is negative two π‘₯ over seven minus three, as before.

So there we have it. Sometimes, when the function is simple, it’s easiest to differentiate it from first principles. That being said, for more complicated functions, the differentiation rules are essential to know.

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