# Video: Differentiating a Combination of Exponential and Rational Functions Using the Chain and the Quotient Rules

Find the derivative of the function π(π§) = β3π^(4π§/(4π§ + 1)).

02:59

### Video Transcript

Find the derivative of the function π of π§ equals negative three π to the four π§ over four π§ plus one.

Here we have a function of a function or a composite function. This tells us weβre going to need to apply the chain rule. This says that the derivative of π of π of π₯ is equal to the derivative of π of π of π₯ multiplied by the derivative of π of π₯. Alternatively, we can say that if π¦ is equal to this composite function, π of π of π₯, then if we let π’ be equal to π of π₯, then π¦ is equal to π of π’. And this means we can say that 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₯.

In this example, we say that π¦ is equal to negative three times π to the power of π’, where π’ is equal to four π§ over four π§ plus one. And since weβre going to be differentiating with respect to π§, we alter the formula slightly. And we say that dπ¦ by dπ§ is equal to dπ¦ by dπ’ times dπ’ by dπ§. So weβre going to need to work out dπ¦ by dπ’ and dπ’ by dπ§. dπ¦ by dπ’ is a fairly easy one to differentiate. We know that the derivative of π to the power of π’ is π to the power of π’. So the derivative of negative three π to the power of π’ is negative three π to the power of π’. And then, we can replace π’ with four π§ over four π§ plus one to get negative three π to the four π§ over four π§ plus one. But what about the derivative of four π§ over four π§ plus one?

Well, here, we need to use the quotient rule. This says that the derivative of π of π₯ over π of π₯ is equal to the function π of π₯ times the derivative of π of π₯ minus the function π of π₯ times the derivative of π of π₯. And thatβs all over π of π₯ squared. We change π of π₯ to π of π§ and π of π₯ to π of π§. Then, the derivative of the numerator of our fraction is four. And the derivative of the denominator of our fraction is also four. So the equivalent to π of π§ times the derivative of π of π§ is four π§ plus one times four. And the equivalent to π of π§ times the derivative of π of π§ is four π§ times four. And thatβs all over the denominator squared. Thatβs four π§ plus one squared. Now, distributing the parentheses and we simply end up with four on the numerator of this fraction. So dπ’ by dπ§ is equal to four over four π§ plus one squared.

Letβs now substitute everything we have into the formula for the chain rule. dπ¦ by dπ’ times dπ’ by dπ§ is negative three π to the four π§ over four π§ plus one times four over four π§ plus one squared. And if we simplify, we see that the derivative of our function is negative 12π to the four π§ over four π§ plus one all over four π§ plus one squared.