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

Find 𝑑𝑦/𝑑π‘₯, given that βˆ’6π‘₯𝑦 = 11.

02:04

Video Transcript

Find 𝑑𝑦 𝑑π‘₯ given that negative six π‘₯𝑦 equals 11.

Now the first thing we want to do before we differentiate is actually rearrange our equation so that 𝑦 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 𝑦 is equal to 11 over negative six π‘₯. Okay, great! But we want to differentiate now. But I think there’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‘Ž to the power of 𝑏 is equal to π‘Ž to the power of negative 𝑏. So great! We now have it in a form that we can differentiate, so let’s go ahead and differentiate our function.

So we’re gonna have 𝑑𝑦 𝑑π‘₯ is equal to β€” and now here’s where we’re gonna use our differentiation rule. And we can say that if we have a function that’s π‘Žπ‘₯ to the power of 𝑏, then if we find the derivative, it’s going to be π‘Žπ‘ β€” so we’re gonna multiply the exponent by the coefficient β€” and then π‘₯ the power of 𝑏 minus one. So you subtract one from the exponent.

Okay, so let’s use this and find 𝑑𝑦 𝑑π‘₯. So we’re gonna get that the derivative is equal to negative 11 over six multiplied by negative one, cause that’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.

I’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 𝑑𝑦 𝑑π‘₯ 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‘Ž to the power of negative 𝑏, then this is equal to one over π‘Ž to the power of 𝑏. So if we have π‘₯ to the power of negative two. This is equal to one over π‘₯ squared.

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