### Video Transcript

Differentiate π· of π‘ is equal to one minus 81π‘ to the fourth power all divided by three π‘ all raised to the fifth power.

Weβre asked to differentiate the function π· of π‘, and we could see that π· of π‘ is the quotient of two functions. In our numerator, we have one minus 81π‘ to the fourth power. And in our denominator, we have three π‘ all raised to the fifth power. So we might be tempted to differentiate this by using the quotient rule. After all, we can differentiate the numerator and the denominator with respect to π‘. And this would work.

However, in this case, thereβs a simpler method. Instead of directly using the quotient rule, first, weβll distribute the exponent of five over the parentheses in our denominator. Doing this, we get that π· of π‘ is equal to one minus 81π‘ to the fourth power all divided by 243π‘ to the fifth power. And now we can see something interesting. We can just divide both terms in our numerator by our denominator. Doing this, we get one over 243π‘ to the fifth power minus 81π‘ to the fourth power divided by 243π‘ to the fifth power.

And now we can start simplifying. In our second term, 81 divided by 243 is just equal to one-third. And we can also cancel four of our shared factors of π‘ in our numerator and our denominator. This gives us one over 243π‘ to the fifth power minus one divided by three π‘. And the last thing weβll do is use our laws of exponents to rewrite our factors of π‘ from the denominator into our numerator. This gives us π‘ to the power of negative five divided by 243 minus π‘ to the power of negative one divided by three.

And now we can see we can evaluate the derivatives of both of these terms by using the power rule for differentiation. We want to multiply by our exponents of π‘ and then reduce this exponent by one, so we can evaluate the derivative of π· of π‘ term by term. In our first term, our exponent of π‘ is negative five. So we need to multiply by this exponent of π‘ and then reduce this exponent by one. This gives us negative five π‘ to the power of negative six divided by 243. And in our second term, our exponent of π‘ is negative one. So we multiply by negative one and then reduce this exponent by one.

So weβre subtracting negative one times π‘ to the power of negative two divided by three. And of course, we can simplify. Subtracting negative one times π‘ to the power of negative two over three is the same as adding π‘ to the power of negative two over three. And finally, weβll rearrange these terms to get one-third π‘ to the power of negative two minus five divided by 243 times π‘ to the power of negative six. And this is our final answer.

In this question, we were given the function π· of π‘ is equal to one minus 81π‘ to the fourth power all divided by three π‘ all raised to the power of five. And we were asked to differentiate this. And instead of doing this by using the quotient rule, we were able to manipulate our expression into a form which we could differentiate by using the power rule for differentiation. We got that π· prime of π‘ is equal to one-third π‘ to the power of negative two minus five divided by 243 π‘ to the power of negative six.