### Video Transcript

If π of π₯ is equal to π₯ squared minus two over two π₯ to the fourth power minus one all cubed, what is the value of π prime of one?

Recall that π prime of one means the first derivative of this function evaluated when π₯ is equal to one. So weβre going to need to find an expression for the derivative of π of π₯. Now looking at π of π₯, we see that it is a relatively complex function. So weβre going to need to find this derivative step by step. We notice first of all that π of π₯ is a composite function. We have the quotient π₯ squared minus two over two π₯ to the fourth power minus one. And then this is all raised to the power of three or cubed. We can find the derivative of a composite function using the chain rule. And to do this, we make a substitution. Weβll introduce the variable π’ to represent the expression inside the parentheses. Thatβs π₯ squared minus two over two π₯ to the fourth power minus one. Then our function π has become a function of π’. Itβs equal to π’ cubed.

The chain rule then tells us that if π is a function of π’ and π’ is a function of π₯, which is what we have here, then the derivative of π with respect to π₯ is equal to the derivative of π with respect to π’ multiplied by the derivative of π’ with respect to π₯. The derivative of π with respect to π’ can be found without too much difficulty. Itβs equal to three π’ squared by applying the power rule of differentiation. However, the derivative of π’ with respect to π₯ is going to be much more complex because we noticed that π’ is a quotient of two differentiable functions. Weβre, therefore, also going to need to apply the quotient rule in order to find the derivative of π’ with respect to π₯. The quotient rule tells us that for two differentiable functions, which here Iβm going to call π and π, the derivative with respect. π₯ of their quotient π over π is equal to π multiplied by dπ by dπ₯ minus π multiplied by dπ by dπ₯ all over π squared.

We will, therefore, let π equal to function in our numerator, π₯ squared minus two, and π equal the function in the denominator, two π₯ to the fourth power minus one. We can then find each of their derivatives with respect to π₯ by applying the power rule of differentiation again. dπ by dπ₯ is equal to two π₯ and dπ by dπ₯ is equal to eight π₯ cubed. We can then substitute into the formula weβve written for the quotient rule. dπ’ by dπ₯ or dπ over π by dπ₯ is equal to π times dπ by dπ₯. Thatβs two π₯ to the fourth power minus one times two π₯. Minus π times dπ by dπ₯. Thatβs π₯ squared minus two times eight π₯ cubed. All over π squared, which is two π₯ to the fourth power minus one all squared.

Distributing the parentheses in the numerator gives four π₯ to the fifth power minus two π₯ minus eight π₯ to the fifth power plus 16π₯ cubed all over two π₯ to the fourth power minus one all squared. And finally, combining the like terms in the numerator to simplify Gives negative four π₯ to the fifth power plus 16π₯ cubed minus two π₯ over two π₯ to the fourth power minus one all squared. And so we have our expression for dπ’ by dπ₯. Now that we know both dπ’ by dπ₯ and dπ by dπ’, we can return to the chain rule. Substituting both derivatives gives dπ by dπ₯ is equal to three π’ squared multiplied by negative four π₯ to the fifth power plus 16π₯ cubed minus two π₯ over two π₯ to the fourth power minus one all squared.

However, notice that this derivatives still has a π’ term. So we need to replace this with its definition in terms of π₯. We reverse our earlier substitution, replacing π’ with π₯ squared minus two over two π₯ to the fourth power minus one. And we have that dπ By dπ₯ is equal to three multiplied by π₯ squared minus two over two π₯ to the fourth power minus one all squared multiplied by negative four π₯ to the fifth power plus 16π₯ cubed minus two π₯ over two π₯ to the fourth power minus one squared. Weβre nearly there. But the final step is we need to evaluate this first derivative when π₯ is equal to one. And actually, thatβs relatively straightforward because one to the power of anything is just one. So everywhere that we see π₯ or π₯ to a power, we can just replace with one.

In fact, what happens when substituting π₯ equals one is each term just reduces to its coefficients. So we have three multiplied by one minus two over two minus one all squared multiplied by negative four plus 16 minus two over two minus one squared. Simplifying gives three multiplied by negative one over one squared multiplied by 10 over one. Negative one over one is just negative one and squaring gives one. So we have three multiplied by one multiplied by 10 which is equal to 30. So by applying both the chain rule and the quotient rule, we were able to find an expression for the first derivative of this complex function. And then by substituting π₯ equals one into our derivative, we found that the value of π prime of one is 30.