# Question Video: ο»Ώ Differentiating Polynomial Functions with Decimal Coefficients Mathematics • Higher Education

Differentiate the function π(π‘) = 6.3π‘β΅ + 5.2π‘Β² + 7.3.

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### Video Transcript

Differentiate the function π of π‘ is equal to 6.3 times π‘ to the fifth power plus 5.2 times π‘ squared plus 7.3.

We need to differentiate the function π of π‘. First, we notice that π is a function of π‘. This means we need to differentiate π with respect to π‘. Next, we notice that our function π of π‘ is a polynomial. This means we can differentiate this term by term by using the power rule for differentiation. Next, we recall the power rule for differentiation tells us, for any real constants π and π, the derivative of π times π‘ to the πth power with respect to π‘ is equal to π times π times π‘ to the power of π minus one. We multiply by the exponent of π‘ and reduce this exponent by one.

Weβre now ready to find our expression for π prime of π‘. Weβll differentiate this term by term. This means, instead of differentiating the entire function π of π‘ with respect to π‘, we can differentiate each term of π of π‘ separately. Letβs start with our first term. We can see the exponent of π‘ is five and our coefficient is 6.3, so we need to multiply by the exponent of π‘, which is five, and then reduce this exponent by one. This gives us five times 6.3π‘ to the power of five minus one. We can do exactly the same in our second term.

This time the coefficient is 5.2 and our exponent of π‘ is equal to two. Once again, we multiply by our exponent of π‘, which is two, and reduce this exponent by one. This gives us two times 5.2π‘ to the power of two minus one. Now, thereβs a few different ways we could evaluate the derivative of our third and final term. For example, we could recall π‘ to the zeroth power is equal to one, so we could rewrite 7.3 as 7.3 times π‘ to the zeroth power and then apply the power rule for differentiation. However, thereβs an easier method. Remember that 7.3 is a constant. It does not change as our value of π‘ changes. Therefore, its rate of change with respect to π‘ is equal to zero. Both of these methods will give us the answer of zero.

We can now start simplifying this expression. First, five times 6.3 is equal to 31.5. And then in our exponent of π‘, five minus one is equal to four. This gives us the first term of 31.5 times π‘ to the fourth power. And we can also simplify our second term. Two times 5.2 is equal to 10.4. And then in our exponent of π‘, two minus one is equal to one. And of course, π‘ to the first power is just equal to π‘, so this gives us the second term of 10.4π‘. And the final term is equal to zero, so we donβt need to include this. Therefore, we were able to show if π of π‘ is equal to 6.3π‘ to the fifth power plus 5.2π‘ squared plus 7.3, then π prime of π‘ is equal to 31.5π‘ to the fourth power plus 10.4π‘.

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