Video: Differentiating Polynomial Functions

Differentiate 𝑓(π‘₯) = (1/2)π‘₯^(18) βˆ’ (1/5)π‘₯^(25) + (1/8)π‘₯^(32).

02:51

Video Transcript

Differentiate 𝑓 of π‘₯ is equal to one-half π‘₯ to the 18th power minus one-fifth π‘₯ to the 25th power plus one-eighth π‘₯ to the 32nd power.

We’re asked to differentiate the function 𝑓 of π‘₯. Since this is a function in π‘₯, we need to differentiate this with respect to π‘₯. And in fact, we can see that 𝑓 of π‘₯ is a polynomial. It doesn’t matter that our exponent of π‘₯ is very large; it’s still just a polynomial. And we know how to find the derivatives of polynomials, we do this term by term by using the power rule for differentiation.

And we recall the power rule for differentiation tells us for any real constants π‘Ž and 𝑛, to differentiate π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯, we multiply by our exponent of π‘₯ and reduce this exponent by one. This gives us π‘›π‘Žπ‘₯ to the power of 𝑛 minus one.

We’re now ready to start finding an expression for 𝑓 prime of π‘₯. We’ll start by writing this derivative out term by term. This gives us the derivative of one-half π‘₯ to the 18th power with respect to π‘₯ plus the derivative of negative one-fifth π‘₯ to the 25th power with respect to π‘₯ plus the derivative of one-eighth π‘₯ to the 32nd power with respect to π‘₯. And now we can evaluate each of these derivatives by using the power rule for differentiation. Let’s start with the first term. Our exponent of π‘₯ is equal to 18, and our coefficient of π‘₯ is equal to one-half.

So to evaluate this derivative, we want to multiply by our exponent of 18 and then reduce this exponent by one. This gives us 18 times one-half multiplied by π‘₯ to the power of 18 minus one. We can now do this with the second term. Our exponent of π‘₯ is equal to 25, and our coefficient of π‘₯ is equal to negative one-fifth.

Once again, we multiply by our exponent of 25 and reduce this exponent by one. This gives us 25 times negative one-fifth multiplied by π‘₯ to the power of 25 minus one. And we do exactly the same with our third and final derivative. Our exponent of π‘₯ is equal to 32, and our coefficient of π‘₯ is equal to one-eighth. And once again, we apply the power rule for differentiation to get 32 times one-eighth multiplied by π‘₯ to the power of 32 minus one.

And now we can start simplifying this. Let’s start by simplifying our first term. We have 18 multiplied by one-half is equal to nine. And then our exponent of 18 minus one can be simplified to give us 17. In our second term, we have 25 multiplied by negative one-fifth is equal to negative five. And we can simplify our exponent to give us 24. Finally, our third and final term can be simplified to give us four π‘₯ to the power of 31.

Therefore, given 𝑓 of π‘₯ is equal to one-half π‘₯ to the 18th power minus one-fifth π‘₯ to the 25th power plus one-eighth π‘₯ to the 32nd power, we were able to show the derivative of 𝑓 with respect to π‘₯ is equal to nine π‘₯ to the 17th power minus five π‘₯ to the 24th power plus four π‘₯ to the 31st power.

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