### Video Transcript

Find the first derivative of the function π¦ equals two π₯ to the power of eight π to the power of five π₯.

Well, we can actually look at our function and see thatβs in the form π¦ equals π’π£. And therefore, what we can actually use is the product rule to actually help us differentiate. And the product rule tells us that ππ¦ ππ₯ equals π’ ππ£ ππ₯ plus π£ ππ’ ππ₯. So therefore, what we need to do is decide in our function, if we want to find the first derivative, what π’ and π£ are going to be.

Well, Iβve actually decided that π’ is gonna be equal to two π₯ to the power of eight. And π£ is equal to π to the power of five π₯. So now what we need to do is actually find ππ’ ππ₯ and ππ£ ππ₯. Iβm gonna start with ππ’ ππ₯. And Iβll do that by differentiating two π₯ to the power of eight, which would just give me 16π₯ to the power of seven. And we get that same way that we differentiate any term. Weβve multiplied the exponent by the coefficient, so eight multiplied by two, which gives us 16. And now weβd reduced the exponent by one, so from eight to seven.

Okay, now we move on to π£ to find ππ£ ππ₯. So now to find ππ£ ππ₯, what weβre actually gonna use is a rule. And that rule states that if π¦ is equal to π to the power of π π₯, then ππ¦ ππ₯ is equal to the derivative of π π₯ multiplied by π to the power of π π₯. And we actually get this from the chain rule, because if we have ππ¦ ππ₯ is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯, then we will have π’ is equal to π π₯ and π¦ is equal to π to the power of π’.

So therefore, weβd have that ππ¦ ππ₯ is equal to the derivative of π π₯ multiplied by π to the power of π’. And thatβs because the derivative of π to the power of π’ is just π to the power of π’ because π is a number for which the gradient of π to the power of π₯ is π to the power of π₯. Okay, so then if we substitute back in for π’, we just get the first derivative of π π₯ multiplied by π to the power of π π₯.

Okay, great! So weβve just seen where weβve actually got this rule from. Letβs use it to actually carry on and differentiate our function. Well, weβre gonna get ππ£ ππ₯ is equal to five multiplied by π to the power of five π₯. And thatβs because the derivative of five π₯ is just five. So weβve got the derivative of π π₯, which was five π₯, multiplied by π to the power of five π₯.

Okay, so now at this stage, we can actually use the product rule to actually find out what ππ¦ ππ₯ or the first root of our function is going to be. So first of all, we have π’, which is two π₯ to the power of eight, multiplied by ππ£ ππ₯, which is five π to the power of five π₯. And this is plus π to the power of five π₯, because thatβs our π£ multiplied by our ππ’ ππ₯, which is 16π₯ to the power of seven.

Okay, so we now are just gonna rearrange this to find our ππ¦ ππ₯ and just tidy up. So therefore, weβre gonna get that the first derivative of our function π¦ equals two π₯ to the power of eight π to the power of five π₯ is equal to 10π₯ to the power of eight multiplied by five π to the power of five π₯ plus 16π₯ to the power of seven multiplied by π to the power of five π₯.