Video: Finding the Derivative of a Composite Function Using the Chain Rule

If 𝑓(π‘₯) = 2𝑒^(cos 5π‘₯), what is the value of 𝑓′(0)?

05:13

Video Transcript

If 𝑓 of π‘₯ equals two 𝑒 to the power of cos five π‘₯, what is the value of 𝑓 prime of zero?

We recall, first of all, that this notation 𝑓 prime of zero means we’re looking to find the first derivative of 𝑓 evaluated at π‘₯ equals zero. So, the first thing we need to do is find a general expression for the first derivative 𝑓 prime of π‘₯. Now, looking at our function 𝑓 of π‘₯, we can see that it’s fairly complicated. It involves an exponential and then this is raised to a trigonometric power, cos of five π‘₯. We’re going to need to recall a number of different rules of differentiation, namely, how to differentiate exponentials, how to differentiate trigonometric terms, but also how to differentiate a composite function; that is a function of another function.

We differentiate composite functions using the chain rule, which tells us that if 𝑓 is a function of 𝑒 and 𝑒 is itself a function of π‘₯, then the derivative of 𝑓 with respect to π‘₯ is equal to the derivative of 𝑓 with respect to 𝑒 multiplied by the derivative of 𝑒 with respect to π‘₯ or d𝑓 by dπ‘₯ equals d𝑓 by d𝑒 times d𝑒 by dπ‘₯. What we’re going to do then is to define the function 𝑒 to be the expression in the exponent. So, 𝑒 is equal to cos of five π‘₯. 𝑓 will then be equal to two 𝑒 to the power of 𝑒. And so, we have 𝑓 as a function of 𝑒 and 𝑒 as a function of π‘₯, which means we can apply the chain rule.

First though, we need to find the individual derivatives d𝑒 by dπ‘₯ and d𝑓 by d𝑒. Beginning with d𝑒 by dπ‘₯, we recall the standard result for differentiating a cosine function, which is that the derivative with respect to π‘₯ of cos of π‘Žπ‘₯ for some constant π‘Ž is equal to negative π‘Ž multiplied by sin of π‘Žπ‘₯ provided the angle is measured in radians. So, we see that d𝑒 by dπ‘₯ will be equal to negative five sin five π‘₯. To find d𝑓 by d𝑒, we need to recall the standard result for differentiating an exponential. And in fact, we remember that the derivative of 𝑒 to the power of π‘₯ is simply 𝑒 to the power of π‘₯. So, the derivative of 𝑒 to the power of 𝑒 with respect to 𝑒 will simply be 𝑒 to the power of 𝑒. And we have a multiplicative constant of two. So, we found each of our individual derivatives d𝑓 by d𝑒 and d𝑒 by dπ‘₯.

The chain rule tells us that to find d𝑓 by dπ‘₯ or 𝑓 prime of π‘₯, we multiply these derivatives together. We have then that 𝑓 prime of π‘₯ is equal to two 𝑒 to the power of 𝑒 multiplied by negative five sin five π‘₯. Now, remember, we’re later going to evaluate this when π‘₯ is equal to zero. But before we can do this, we need to make sure our derivative is in terms of π‘₯ only. So, we need to undo the substitution that we made. We define 𝑒 to be equal to cos of five π‘₯. So replacing 𝑒 with this expression in π‘₯ and simplifying the coefficient at the same time, we now have that 𝑓 prime of π‘₯ is equal to negative 10𝑒 to the power of cos five π‘₯ sin five π‘₯. To evaluate this derivative for a particular π‘₯-value, we just need to substitute that π‘₯-value throughout.

So, we have that 𝑓 prime of zero is a equal to negative 10 multiplied by 𝑒 to the power of cos of five times zero multiplied by sin of five times zero. Of course, five times zero is simply zero. And we then recall that cos of zero is one and sin of zero is zero. So, we have negative 10𝑒 to the first power multiplied by zero. But of course, multiplying anything by zero simply results in an answer of zero. By applying the chain rule then and recalling standard results for differentiating trigonometric and exponential functions, we found that for this function 𝑓 of π‘₯, its first derivative evaluated at π‘₯ equals zero is simply zero.

Now, in fact, there is a slight shortcut we could have taken if we recall a more general result for differentiating exponential functions. We recall that if we have an exponential to the power of some function 𝑔 of π‘₯, then its derivative with respect to π‘₯ is equal to 𝑔 prime of π‘₯ multiplied by 𝑒 to the power of 𝑔 of π‘₯. In this case then, our function 𝑔 of π‘₯ is the exponent. That’s cos of five π‘₯. Its derivative with respect to π‘₯ we already know it’s negative five multiplied by sin five π‘₯.

So by applying this standard result for differentiating exponential functions, we see that 𝑓 prime of π‘₯ is equal to two multiplied by negative five sin five π‘₯ multiplied by 𝑒 to the power of cos five π‘₯. Simplifying and we get negative 10𝑒 to the power of cos five π‘₯ multiplied by sin five π‘₯, which we see is what we had for our derivative at this stage here.

So, it’s a slight shortcut. But it’s really just a slightly more informal version of the chain rule. Whichever method you use, we get the same answer. We’ve found that 𝑓 prime of zero is equal to zero.

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