Question Video: Finding the First Derivative of a Function Involving Negative Exponents Using the Power Rule | Nagwa Question Video: Finding the First Derivative of a Function Involving Negative Exponents Using the Power Rule | Nagwa

Question Video: Finding the First Derivative of a Function Involving Negative Exponents Using the Power Rule Mathematics • Second Year of Secondary School

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Find ππ¦/ππ₯, given that π¦ = β43/π₯βΈ.

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Video Transcript

Find ππ¦ ππ₯, given that π¦ equals negative 43 over π₯ to the power of eight.

So to find ππ¦ ππ₯, what weβre gonna need to do is actually differentiate our function. So the first thing Iβm gonna do is Iβm actually going to rewrite our function. And in order to do that, what Iβm gonna use is an exponent rule. And the exponent rule is that one over π to the power of π is equal to π to the power of negative π. So therefore, we can say that π¦ is equal to negative 43π₯ to the power of negative eight. And Iβve done this so that we can actually remove the fraction and it makes it easier to differentiate.

Okay, now, weβre actually gonna move on to the differentiation. And if we want to differentiate, we think about our function in the form ππ₯ to the power of π. So if we have a function in this form, then we can say that the derivative which Iβve denoted here is π ππ₯ of π π₯ is gonna be equal to ππ π₯ to the power of π minus one. So what Iβve done here is Iβve actually multiplied our coefficient by our exponents β so π and π. And then, weβve reduced the exponent by one. So weβve got π minus one.

Okay, great, we know what to do. Itβs to use this to actually differentiate our function. So then, we can say that the derivative of our function or ππ¦ ππ₯ is equal to negative 43 multiplied by negative eight cause thatβs our coefficient multiplied by our exponent and then π₯ to the power of negative eight minus one. Right, so now, we can actually simplify this and this gives us 344π₯ to the power of negative nine. And we got 344 because we got negative 43 multiplied by negative eight. And a negative multiplied by a negative gives us a positive. And weβve got π₯ to the power of negative nine because negative eight minus one gives us negative nine.

So fantastic, weβve actually reached an answer here. Weβve differentiated our term. The final thing weβre gonna do is actually rewrite it in our original form β so include a fraction again. And in order to do this, what Iβll use is the exponent rule that I used earlier just in the opposite way. And that rule was the one over π to the power of π equals π to the power of negative π. So we can say that given that π¦ equals negative 43 over π₯ to the power of eight ππ¦ ππ₯ is equal to 344 over π₯ to the power of nine.

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