# Video: Using the Chain Rule and the Quotient Rule to Differentiate a Function

If 𝑓(𝑥) = ((𝑥² − 2)/(2𝑥⁴ − 1))³, what is the value of 𝑓′(1)?

05:30

### Video Transcript

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

Recall that 𝑓 prime of one means the first derivative of this function evaluated when 𝑥 is equal to one. So we’re going to need to find an expression for the derivative of 𝑓 of 𝑥. Now looking at 𝑓 of 𝑥, we see that it is a relatively complex function. So we’re going to need to find this derivative step by step. We notice first of all that 𝑓 of 𝑥 is a composite function. We have the quotient 𝑥 squared minus two over two 𝑥 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. We’ll introduce the variable 𝑢 to represent the expression inside the parentheses. That’s 𝑥 squared minus two over two 𝑥 to the fourth power minus one. Then our function 𝑓 has become a function of 𝑢. It’s equal to 𝑢 cubed.

The chain rule then tells us that if 𝑓 is a function of 𝑢 and 𝑢 is a function of 𝑥, which is what we have here, then the derivative of 𝑓 with respect to 𝑥 is equal to the derivative of 𝑓 with respect to 𝑢 multiplied by the derivative of 𝑢 with respect to 𝑥. The derivative of 𝑓 with respect to 𝑢 can be found without too much difficulty. It’s equal to three 𝑢 squared by applying the power rule of differentiation. However, the derivative of 𝑢 with respect to 𝑥 is going to be much more complex because we noticed that 𝑢 is a quotient of two differentiable functions. We’re, therefore, also going to need to apply the quotient rule in order to find the derivative of 𝑢 with respect to 𝑥. The quotient rule tells us that for two differentiable functions, which here I’m going to call 𝑝 and 𝑞, the derivative with respect. 𝑥 of their quotient 𝑝 over 𝑞 is equal to 𝑞 multiplied by d𝑝 by d𝑥 minus 𝑝 multiplied by d𝑞 by d𝑥 all over 𝑞 squared.

We will, therefore, let 𝑝 equal to function in our numerator, 𝑥 squared minus two, and 𝑞 equal the function in the denominator, two 𝑥 to the fourth power minus one. We can then find each of their derivatives with respect to 𝑥 by applying the power rule of differentiation again. d𝑝 by d𝑥 is equal to two 𝑥 and d𝑞 by d𝑥 is equal to eight 𝑥 cubed. We can then substitute into the formula we’ve written for the quotient rule. d𝑢 by d𝑥 or d𝑝 over 𝑞 by d𝑥 is equal to 𝑞 times d𝑝 by d𝑥. That’s two 𝑥 to the fourth power minus one times two 𝑥. Minus 𝑝 times d𝑞 by d𝑥. That’s 𝑥 squared minus two times eight 𝑥 cubed. All over 𝑞 squared, which is two 𝑥 to the fourth power minus one all squared.

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

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

In fact, what happens when substituting 𝑥 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 𝑥 equals one into our derivative, we found that the value of 𝑓 prime of one is 30.