### Video Transcript

Find the first derivative of the function π¦ equals two π to the power of four π₯ minus five π to the power of negative five π₯ all to the fourth power.

Here, we have a function of a function, sometimes called a composite function. And weβre looking to find the first derivative. That is, weβre going to differentiate π¦ with respect to π₯ exactly once. Because weβre dealing with a composite function though, weβre going to need to use the chain rule to do so. The chain rule says that if π¦ is some function in π’, and π’ itself is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.

So, weβre going to need to define our function π’. Here, weβre going to let π’ be the inside function. Thatβs two π to the power of four π₯ minus five π to the power of negative five π₯. And that means we can rewrite π¦ as π¦ equals π’ to the fourth power. Weβre going to need to work out what dπ’ by dπ₯ is and what dπ¦ by dπ’ is. Weβll start by finding dπ¦ by dπ’. Thatβs the derivative of our function π¦ with respect to π’ . And here, we recall the rule for differentiating a function in the form ππ₯ to the power of π for nonzero real constants π and π.

That is π times ππ₯ to the power of π minus one. Essentially, we multiply the entire function by the power and then reduce that power by one. So, dπ¦ by dπ’ is four times π’ to the power of four minus one. Thatβs four π’ cubed. But how do we differentiate our function π’ with respect to π₯? Well, we could recall the fact that the derivative of the function π to the power of π₯ is π to the power of π₯ and then use the chain rule to work out the derivative of π the power of four π₯.

However, itβs absolutely fine to quote the standard result that the derivative of π to the power of ππ₯ is π times π to the power of ππ₯. We also know that to find the derivative of a constant multiplied by a function, we can multiply that constant by the derivative of the function itself. So, we can differentiate π to the power of four π₯ to get four π to the power of four π₯. And then, we can multiply that by two. So, thatβs two times four π to the power of four π₯. Similarly, when we differentiate π to the power of negative five π₯, we get negative five π to the power of negative five π₯.

So, dπ’ by dπ₯ is two times four π to the power of four π₯ minus five times negative five π to the power of negative five π₯. And we obtain dπ’ by dπ₯ to be equal to eight π to the power of four π₯ plus 25π to the power of negative five π₯. We can now substitute everything we have into our formula for the chain rule. When we do, we see that dπ¦ by dπ₯ is equal to four π’ cubed multiplied by eight π to the power of four π₯ plus 25π to the power of negative five π₯.

We do have a bit of a problem though. Weβre wanting to differentiate π¦ with respect to π₯. And currently we have an expression for the derivative in terms of π’ and π₯. So, we go back to our definition of π’. We said that π’ was equal to two π to the power four π₯ minus five π to the power of negative five π₯. So, letβs replace π’ in our expression for the derivative with this function. And it becomes four times two π to the power of four π₯ minus five times π to the power of negative five π₯ cubed times eight π to the power of four π₯ plus 25π to the power of negative five π₯. And so, by using the chain rule and quoting some standard results for the derivative of polynomial terms and exponential functions, we found dπ¦ by dπ₯.