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

Find 𝑑𝑦/𝑑π‘₯, given that 𝑦 = 3π‘₯⁴ + 4π‘₯Β² + 6 βˆ’ (7/π‘₯⁷) βˆ’ (8/π‘₯⁸).

04:25

Video Transcript

Find 𝑑𝑦 𝑑π‘₯, given that 𝑦 equals three π‘₯ to the power of four plus four π‘₯ squared plus six minus seven over π‘₯ to the power of seven minus eight over π‘₯ to the power of eight.

Now, in order to actually find 𝑑𝑦 𝑑π‘₯, what we’re gonna have to do is actually differentiate our function. However, before I do that, what I want to actually do is rewrite it in a way such that we remove the fractions. And to do this, what we’re gonna use is the exponent rule. And that rule tells us that if we have one over π‘Ž to the power of 𝑏, then this is gonna be equal to π‘Ž to the power of negative 𝑏.

Okay, great, so let’s use this and actually rewrite our function. So now, we’ve actually used this. We’ve got 𝑦 is equal to three π‘₯ to the power of four plus four π‘₯ squared plus six minus seven π‘₯ to the power of negative seven minus eight π‘₯ to the power of negative eight. And the way in which we actually use our exponent rule to get these last two terms was by β€” if you wanted to think of it this way β€” we had seven multiplied by one over π‘₯ to the power of seven. So if I split it up like that, so therefore we’d actually get seven multiplied by π‘₯ to the power of negative seven. So we get seven π‘₯ to the power of negative seven.

Okay, now that we have this, we can actually get on and differentiate our function. Now in order to actually differentiate our function, we actually use something called the general power rule. And this tells us that if we have a function in the form π‘Žπ‘₯ to the power of 𝑏, then the derivative of this function is gonna be equal to π‘Žπ‘π‘₯ to the power of 𝑏 minus one. So what we’ve actually done in practice is we’ve multiplied the exponent by the coefficient β€” so π‘Žπ‘ β€” and then we’ve actually reduced the exponent by one β€” so 𝑏 minus one.

Okay, so now what we can do is we can actually use this to differentiate our function. So then using a differentiation rule, we have our first term which is 12π‘₯ to the power of three. So the way we actually differentiated three π‘₯ to the power of four was we multiplied our coefficient by our exponent β€” so three by four β€” which gave us 12 and then it’s π‘₯ to the power of four minus one cause we subtract one from the exponent which gave us π‘₯ to the power of three. And then, our next term is plus eight π‘₯ and we got that in the same way we did four multiplied by two, coefficient multiplied by exponent, gives us eight. And then, we had π‘₯ to the power of two minus one gives us just π‘₯ or π‘₯ to the power of one.

Then, looking at our next term, this would just be plus zero which we wouldn’t usually write, but I’m just putting in here so we can actually think about how we got that. Well, if we actually think about it like this, we’ve got the derivative of six π‘₯ to the power of zero. And I say that because if we have six, then it’s just six multiplied by one. Well, π‘₯ to the power of zero is one.

So how would we actually differentiate that? Well, what we’d have is the coefficient of six multiplied by the exponent of zero and π‘₯ to the power of zero minus one. Well, zero multiplied by anything is just zero. So therefore, our term is going to be just zero.

Okay, great, let’s move on to the last two terms. Well, our next term is gonna be positive 49π‘₯ to the power of negative eight. Well, just to draw our attention to how we actually did this one just because it’s got a couple of negatives in there, we’ve got the derivative of negative seven π‘₯ to the power of negative seven. Well, then we have a negative seven because that’s our coefficient multiplied by negative seven which is our exponent. So therefore, negative multiplied by negative gives us a positive and that’s 49. And then, we have π‘₯ to the power of negative seven minus one, which gives us negative eight.

And then, our final term works in exactly the same way and we get positive 64π‘₯ to the minus nine because we had negative eight multiplied by negative eight which gives us positive 64. And then, we had π‘₯ to the power of negative eight minus one which gives us π‘₯ to the power of negative nine. Okay, fantastic, we’ve actually got each of our terms. Now, let’s actually simplify and put it back into its original form.

So therefore, we can say that given that 𝑦 equals three π‘₯ to the power of four plus four π‘₯ squared plus six minus seven over π‘₯ to the power of seven minus eight over π‘₯ to the power of eight, then 𝑑𝑦 𝑑π‘₯ is gonna be equal to 12π‘₯ cubed plus eight π‘₯ plus 49 over π‘₯ to the power of eight plus 64 over π‘₯ to the power of nine. And we got the last two terms in that form just by using our exponent rule that we used at the beginning, which was that one over π‘Ž to the power of 𝑏 is equal to π‘Ž to the power of negative 𝑏.

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