Video: Differentiating Compositions of Exponential and Root Functions Using the Chain Rule

Find the derivative of the function 𝑦 = √(3π‘₯𝑒^(βˆ’4π‘₯) + 4).

05:58

Video Transcript

Find the derivative of the function 𝑦 is equal to the square root of three π‘₯ times 𝑒 to the power of negative four π‘₯ plus four.

We’re given 𝑦 which is a composition of many functions. We need to find the derivative of 𝑦. In fact, since 𝑦 is a function in π‘₯, we need to find the derivative of 𝑦 with respect to π‘₯. To start this question, we’ll need to look at what type of function 𝑦 is. We can see inside of the square root sign, we have the product of two functions three π‘₯ multiplied by 𝑒 to the power of negative four π‘₯ and then we add four. So we’re taking the square root of the product of two functions. This means we’re going to need to use the product rule and the chain rule.

However, we could also use our laws of exponents to rewrite 𝑦 as three π‘₯ times 𝑒 to the power of negative four π‘₯ plus four all raised to the power of one-half. And now, we can differentiate this by using both the product rule and the general power rule. We can do either of these two methods. In this video, we’ll use the general power rule.

To start, let’s call 𝑓 of π‘₯ three π‘₯ times 𝑒 to the power of negative four π‘₯ plus four. That’s the entire function inside the square root symbol of 𝑦. So by using this definition for our function 𝑓 of π‘₯, we’ve rewritten 𝑦 as the square root of 𝑓 of π‘₯, or by using our laws of exponents, 𝑓 of π‘₯ all raised to the power of one-half. And of course, we know that our function 𝑓 of π‘₯ is differentiable. This is because it’s the product of differentiable functions. So to find the derivative of 𝑦 with respect to π‘₯, we’ll use the general power rule.

So let’s recall what the general power rule tells us. The general power rule states for a differentiable function 𝑓 of π‘₯ and real constant 𝑛, the derivative of 𝑓 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ all raised to the power of 𝑛 minus one.

In our case, we have the function 𝑓 of π‘₯ which we know is differentiable. And our exponent value of 𝑛 will be equal to one-half. So now, we can apply the general power rule to find the derivative of 𝑦 with respect to π‘₯. We get 𝑦 prime is equal to one-half times 𝑓 prime of π‘₯ multiplied by three π‘₯ times 𝑒 to the power of negative four π‘₯ plus four all raised to the power of one-half minus one. And we can simplify this slightly. In our exponent, one-half minus one is equal to negative one-half.

But we’re not done. To find an expression for 𝑦 prime, we now need to find an expression for 𝑓 prime of π‘₯. And if we look at our definition of the function 𝑓 of π‘₯, to do this, we’re going to need to use the product rule. So let’s now recall the product rule.

The product rule tells us if a function 𝑓 of π‘₯ is the product of two differentiable functions 𝑒 of π‘₯ times 𝑣 of π‘₯, then 𝑓 prime of π‘₯ is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ plus 𝑣 prime of π‘₯ times 𝑒 of π‘₯. And we can use this to find an expression for 𝑓 prime of π‘₯. That’s the derivative of three π‘₯ times 𝑒 to the power of negative four π‘₯ plus four with respect to π‘₯.

First, we’ll evaluate this derivative term by term. That’s the derivative of three π‘₯ times 𝑒 to the power of negative four π‘₯ with respect to π‘₯ plus the derivative of four with respect to π‘₯. Of course, four is a constant, so its derivative with respect to π‘₯ is equal to zero. So in actual fact, to find an expression for 𝑓 prime of π‘₯, we only need to differentiate three π‘₯ times 𝑒 to the power of negative four π‘₯. So we’ll set our function 𝑒 of π‘₯ to be three π‘₯ and 𝑣 of π‘₯ to be 𝑒 to the power of negative four π‘₯.

To apply the product rule, we’re going to need to find expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯. Let’s start with 𝑒 prime of π‘₯. That’s the derivative of three π‘₯ with respect to π‘₯. Three π‘₯ is a linear function, so its derivative with respect to π‘₯ will be the coefficient of π‘₯, which is three.

Let’s now find an expression for 𝑣 prime of π‘₯. That’s the derivative of 𝑒 to the power of negative four π‘₯ with respect to π‘₯. To do this, we’re going to need to recall one of our standard rules for differentiating exponential functions. For any real constant π‘Ž, the derivative of 𝑒 to the power of π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times 𝑒 to the power of π‘Žπ‘₯.

In our case, our value of π‘Ž is equal to negative four. So we get 𝑣 prime of π‘₯ is equal to negative four times 𝑒 to the power of negative four π‘₯. We’re now ready to use the product rule to find an expression for 𝑓 prime of π‘₯. In fact, we could’ve just slightly changed our product rule to evaluate this from the first line of our expression. So the product rule tells us 𝑓 prime of π‘₯ will be equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ plus 𝑣 prime of π‘₯ times 𝑒 of π‘₯.

Substituting in our expressions for 𝑒 of π‘₯, 𝑣 of π‘₯, 𝑒 prime of π‘₯, and 𝑣 prime of π‘₯, we get that 𝑓 prime of π‘₯ is equal to three times 𝑒 to the power of negative four π‘₯ plus negative four 𝑒 to the power of negative four π‘₯ multiplied by three π‘₯. And we’ll simplify this expression by taking out the shared factor of three times 𝑒 to the power of negative four π‘₯. This gives us three 𝑒 to the power of negative four π‘₯ multiplied by one minus four π‘₯.

But remember, the question is asking us to find an expression for the derivative of 𝑦 with respect to π‘₯. So we need to substitute this expression for 𝑓 prime of π‘₯ into our expression for 𝑦 prime. Substituting in our expression for 𝑓 prime of π‘₯, we get 𝑦 prime is equal to one-half multiplied by three 𝑒 to the power of negative four π‘₯ times one minus four π‘₯ multiplied by three π‘₯ times 𝑒 to the power of negative four π‘₯ plus four all raised to the power of negative one-half.

And we could leave our answer like this. However, there’s a little bit of simplification we can do. First, by using our laws of exponents, π‘Ž raised to the power of negative one-half is equal to one divided by the square root of π‘Ž. So we’ll use this to move three π‘₯𝑒 to the power of negative four π‘₯ plus four all raised to the power of negative one-half into our denominator. This gives us three times 𝑒 to the power of negative four π‘₯ multiplied by one minus four π‘₯ all divided by two times the square root of three π‘₯𝑒 to the power of negative four π‘₯ plus four.

And the last piece of simplification we’ll do is rearrange our terms one and negative four π‘₯. And this gives us our final answer. Therefore, given the function 𝑦 is equal to the square root of three π‘₯𝑒 to the power of negative four π‘₯ plus four, we were able to use the product rule and the general power rule to find an expression for the derivative of 𝑦 with respect to π‘₯. We got this was equal to three times 𝑒 to the power of negative four π‘₯ multiplied by negative four π‘₯ plus one all divided by two times the square root of three π‘₯𝑒 to the power of negative four π‘₯ plus four.

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