Video: Finding a Limit of a Function Which Can Be Interpreted as the Differential of Another Function

Find lim_(β„Ž β†’ 0) (((π‘₯ + β„Ž)⁷ cos ((π‘₯ + β„Ž)Β³) βˆ’ (π‘₯⁷ cos (π‘₯Β³)))/β„Ž)

03:16

Video Transcript

Find the limit as β„Ž approaches zero of π‘₯ plus β„Ž to the seventh power times cos of π‘₯ plus β„Ž cubed minus π‘₯ to the seventh power times cos of π‘₯ cubed all over β„Ž.

Now, what we’re not going to do here is to find a way to manipulate the expression in our limit. Instead, we’re going to recognize what this limit actually represents. Remember, the derivative of a function 𝑓 at a given point π‘Ž denoted by 𝑓 prime of π‘Ž is the limit as β„Ž approaches zero of a 𝑓 of π‘Ž plus β„Ž minus 𝑓of π‘Ž all over β„Ž. Now more generally, if we replace π‘Ž with π‘₯, we find that the derivative 𝑓 prime of π‘₯ is equal to the limit as β„Ž approaches zero of 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ all over β„Ž.

Comparing the general form of the derivative to our limit and we see that 𝑓 of π‘₯ must be equal to this function. 𝑓 of π‘₯ is equal to π‘₯ to the seventh power times cos of π‘₯ cubed. And we know that our limit then is equal to the derivative of this function 𝑓 prime of π‘₯. So rather than evaluating the limit, let’s evaluate the derivative. Now, this function π‘₯ to the seventh power times cos of π‘₯ cubed is itself the product of two differentiable functions. And so we’re going to use the product rule to find its derivative. This says that the derivative of the product to do differentiable functions is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯.

Let’s say that 𝑒 is equal to π‘₯ to the seventh power and 𝑣 is equal to cos of π‘₯ cubed. According to the product rule, we’re going to need to differentiate each of these with respect to π‘₯. Well, the derivative of π‘₯ to the seventh power is quite straightforward. We multiply by the entire exponent and then reduce that exponent by one. So d𝑒 by dπ‘₯, the derivative of 𝑒 with respect to π‘₯, is seven π‘₯ to the sixth power. d𝑣 by dπ‘₯ is a little bit more tricky since we have a composite function. And so we’re going to use the chain rule. This says that if 𝑦 is some function in 𝑒 and 𝑒 itself is some function in π‘₯, then the derivative of 𝑦 with respect to π‘₯ is d𝑦 by d𝑒 times d𝑒 by dπ‘₯.

Now, in this case, 𝑦 is the function in 𝑒. So let’s call that cos of 𝑒, where 𝑒 is π‘₯ cubed. The derivative of 𝑦 with respect to 𝑒, the derivative of cos of 𝑒 with respect to 𝑒, is negative sin 𝑒. And then the derivative of 𝑒 with respect to π‘₯ is three π‘₯ squared. Replacing 𝑒 with π‘₯ cubed and we find that d𝑣 by dπ‘₯ is negative three π‘₯ squared sin of π‘₯ cubed. Then the derivative of 𝑒 times 𝑣 β€” in other words, 𝑓 prime of π‘₯ β€” is 𝑒 times d𝑣 by dπ‘₯. So that’s π‘₯ to the seventh power times negative three π‘₯ squared times sin π‘₯ cubed plus 𝑣 times d𝑒 by dπ‘₯. That’s cos of π‘₯ cubed times seven π‘₯ to the sixth power.

Let’s rewrite this as seven π‘₯ to the sixth power times cos of π‘₯ cubed minus three π‘₯ to the ninth power times sin of π‘₯ cubed. Then we can take out a constant factor of π‘₯ of the sixth power. And this means that 𝑓 prime of π‘₯ is π‘₯ to the sixth power times seven cos of π‘₯ cubed minus three π‘₯ cubed times sin of π‘₯ cubed.

Now, if you’re wondering where the three π‘₯ cubed comes from, remember, π‘₯ to the ninth power divided by π‘₯ to the sixth power is π‘₯ cubed since we subtract the exponents. And therefore, three π‘₯ to the ninth power divided by π‘₯ to the sixth power is three π‘₯ cubed. And so whilst we’ve evaluated 𝑓 prime of π‘₯, we also know that this is equal to the limit as β„Ž approaches zero of π‘₯ plus β„Ž to the seventh power times cos of π‘₯ plus β„Ž cubed minus π‘₯ to the seventh power times cos of π‘₯ cubed over β„Ž. It’s π‘₯ to the sixth power times seven cos of π‘₯ cubed minus three π‘₯ cubed time sin of π‘₯ cubed.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.