Video: Differentiating Logarithmic Functions

Find 𝑑𝑦/𝑑π‘₯, given that 𝑦 = 6 log₆ 6π‘₯.

03:39

Video Transcript

Find 𝑑𝑦 by 𝑑π‘₯, given that 𝑦 equals six log base six of six π‘₯.

We need to differentiate this expression involving the function that log base six. But the only logarithmic function that we know how to differentiate is the logarithm base 𝑒 β€” the natural logarithm. 𝑑 by 𝑑π‘₯ of the natural logarithm of π‘₯ is one over π‘₯. It’s not immediately obvious how to differentiate the logarithm base six. But if we can somehow rewrite the expression using the natural logarithm, then we know how to differentiate that.

Luckily, we have the change of base formula: the logarithm base 𝑏 of π‘₯ is the logarithm base π‘Ž of π‘₯ divided by the logarithm base π‘Ž of 𝑏. And choosing π‘Ž to be 𝑒, we can write log base 𝑏 of π‘₯ in terms of the logarithm base 𝑒 β€” in other words the natural logarithm. Instead of writing log base 𝑒, we can use ln to remind us that this is the natural logarithm.

And we can also express the rule using 𝑐 instead of π‘₯, it’s the same rule here but all this save us some confusion when we come to substitute. We have a log base six of six π‘₯. So comparing this to log base 𝑏 of 𝑐, we see that 𝑏 is six and 𝑐 is six π‘₯. And so log base six of six π‘₯ is the natural logarithm of six π‘₯ over the natural logarithm of six.

Now, we can use the fact that the derivative of a number times a function is that number times the derivative of the function to pull out some constants from inside β€” our 𝑑 by 𝑑π‘₯. We then just have to find the derivative of the natural logarithm of six π‘₯ with respect to π‘₯. And this is a straightforward application of the chain rule. If we let 𝑧 equals six π‘₯ and then apply the chain rule, we’ll get this 𝑑 by 𝑑𝑧 of the natural logarithm of 𝑧 times 𝑑𝑧 by 𝑑π‘₯, where 𝑧 is six π‘₯ as mentioned before.

You might like to pause the video and check that this is indeed what you get when you apply the chain rule. What is 𝑑 by 𝑑𝑧 of the natural logarithm of 𝑧? The derivative with respect to π‘₯ of the natural logarithm of π‘₯ is one over π‘₯ and so the derivative with respect to 𝑧 of the natural logarithm of 𝑧 is one over 𝑧. And 𝑧 is six π‘₯. So we can write this as one over six π‘₯.

How about 𝑑𝑧 by 𝑑π‘₯? Well, 𝑧 is six π‘₯. So differentiating with respect to π‘₯, we get six. Now, we can simplify inside the parentheses. One over six π‘₯ times six is just one over π‘₯. We have shown that the derivative of the natural logarithm of six π‘₯ with respect to π‘₯ is one over π‘₯. And of course, this is also the derivative with respect to π‘₯ of the natural logarithm of just π‘₯.

You might like to think about why the derivative of ln six π‘₯ is the same as the derivative of ln π‘₯. I promise you that this isn’t just an arithmetic error. This really is true. Anyway, back to the problem at hand, the only thing left to do is to write this product’s fractions as a single fraction. Doing so, we find that 𝑑𝑦 by 𝑑π‘₯ is six over π‘₯ ln six.

The thing to take away from this video is that if you want to differentiate some expression which involves the logarithm to some base, then first use a change of base formula to rewrite the expression in terms of the natural logarithm because we know how to differentiate that.

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