Video: Differentiating Polynomial Functions

Differentiate 𝑓(π‘₯) = 8π‘₯Β² βˆ’ 7π‘₯ βˆ’ 2.

03:10

Video Transcript

Differentiate 𝑓 of π‘₯ is equal to eight π‘₯ squared minus seven π‘₯ minus two.

We’re being asked to differentiate a quadratic function. To do this, we first need to remember we can differentiate a function term by term. In other words, if 𝑓 of π‘₯ is equal to 𝑔 of π‘₯ plus β„Ž of π‘₯, then instead of calculating 𝑓 prime of π‘₯, we can instead find 𝑔 prime of π‘₯ and β„Ž prime of π‘₯ and add these two together. And this is also true for the difference between two functions. And in fact, it doesn’t matter how many functions we’re adding and subtracting for each other. We can do this regardless.

So instead of differentiating 𝑓 of π‘₯ all at once, we’ll differentiate it term by term. But to do this, we’ll need to remember the power rule for differentiation. We recall the power rule for differentiation tells us for any real constants π‘Ž and 𝑛, the derivative of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘›π‘Ž times π‘₯ to the power of 𝑛 minus one. We multiply by the exponent of π‘₯ and then reduce this exponent by one. We’ll use this to differentiate 𝑓 of π‘₯ term by term.

Let’s start with the first term, eight π‘₯ squared. We can see that our exponent of π‘₯ is equal to two and our value of π‘Ž is equal to eight. So we want to multiply by this exponent of two and then reduce this exponent by one. This gives us two times eight π‘₯ to the power of two minus one. And of course, we can simplify this to give us 16π‘₯ to the first power, and π‘₯ to the first power is just equal to π‘₯. So the derivative of eight π‘₯ squared with respect to π‘₯ is equal to 16π‘₯.

We’ll now move on to our second term, seven π‘₯. And remember, since we did not include the negative symbol, we will need to subtract this from the derivative of eight π‘₯ squared. To differentiate this, we need to recall that π‘₯ to the first power is equal to π‘₯. So now we can see our exponent of π‘₯ is equal to one and our value of π‘Ž is equal to seven. Once again, we multiply by our exponent of π‘₯ and then reduce this exponent by one. This gives us one times seven π‘₯ to the power of one minus one. And we can simplify this to give us seven π‘₯ to the zeroth power. But π‘₯ to the zeroth power is just equal to one. So this just simplifies to give us seven.

Finally, we need to find the derivative of our third term, two. And once again, we’ll need to subtract this from the rest of our terms. This time, to differentiate this term, we need to recall π‘₯ to the zeroth power is equal to one. So we can write two as two times π‘₯ to the zeroth power. Applying the power rule for differentiation, we get zero times two times π‘₯ to the power of zero minus one. But we can see we’re multiplying by zero. So this in fact just simplifies to give us zero. And in actual fact, the derivative of any constant will always be equal to zero.

We’re now ready to use these to find the expression for 𝑓 prime of π‘₯. It’s equal to the derivative of eight π‘₯ squared minus seven π‘₯ minus two with respect to π‘₯. But we know we can just evaluate the derivative term by term. This gives us the derivative of eight π‘₯ squared with respect to π‘₯ minus the derivative of seven π‘₯ with respect to π‘₯ minus the derivative of two with respect π‘₯. And we already found the derivatives of each of these terms. We get 16π‘₯ minus seven. Therefore, we were able to show if 𝑓 of π‘₯ is equal to eight π‘₯ squared minus seven π‘₯ minus two, then 𝑓 prime of π‘₯ is equal to 16π‘₯ minus seven.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.