Question Video: Finding the First Derivative of a Reciprocal Function Using the Power Rule | Nagwa Question Video: Finding the First Derivative of a Reciprocal Function Using the Power Rule | Nagwa

# Question Video: Finding the First Derivative of a Reciprocal Function Using the Power Rule Mathematics • Second Year of Secondary School

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Find ππ¦/ππ₯, given that β6π₯π¦ = 11.

02:04

### Video Transcript

Find ππ¦ ππ₯ given that negative six π₯π¦ equals 11.

Now the first thing we want to do before we differentiate is actually rearrange our equation so that π¦ is the subject. And in order to this, what we first have to do is actually divide each side by negative six π₯. And this will give us π¦ is equal to 11 over negative six π₯. Okay, great! But we want to differentiate now. But I think thereβs still one more step we should do, and that step is actually to rewrite it in index form.

And that will give us negative 11 over six π₯ to the power of negative one. And we actually got that because we used an exponent rule, and the exponent rule we used is that one over π to the power of π is equal to π to the power of negative π. So great! We now have it in a form that we can differentiate, so letβs go ahead and differentiate our function.

So weβre gonna have ππ¦ ππ₯ is equal to β and now hereβs where weβre gonna use our differentiation rule. And we can say that if we have a function thatβs ππ₯ to the power of π, then if we find the derivative, itβs going to be ππ β so weβre gonna multiply the exponent by the coefficient β and then π₯ the power of π minus one. So you subtract one from the exponent.

Okay, so letβs use this and find ππ¦ ππ₯. So weβre gonna get that the derivative is equal to negative 11 over six multiplied by negative one, cause thatβs coefficient multiplied by exponent, and then π₯ the power of negative one minus one, cause we subtract one from the exponent. And this is gonna give us 11 over six π₯ to the power of negative two.

Iβm gonna do one more stage cause actually we want to do is actually move it back out of index form. So therefore, we can say that ππ¦ ππ₯ is equal to 11 over six π₯ squared. And we actually got to that point because we reversed the rule we used earlier that said that if we have π to the power of negative π, then this is equal to one over π to the power of π. So if we have π₯ to the power of negative two. This is equal to one over π₯ squared.

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