Question Video: Finding the First Derivative of an Exponential Function with an Integer Base | Nagwa Question Video: Finding the First Derivative of an Exponential Function with an Integer Base | Nagwa

Question Video: Finding the First Derivative of an Exponential Function with an Integer Base Mathematics

If 𝑦 = βˆ’3 Γ— 2^π‘₯, determine 𝑑𝑦/𝑑π‘₯.

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Video Transcript

If 𝑦 equals negative three times two to the π‘₯, determine 𝑑𝑦 by 𝑑π‘₯.

We want to find 𝑑𝑦 by 𝑑π‘₯. And as 𝑦 equals negative three times two to the π‘₯, that means differentiating negative three times two to the π‘₯ with respect to π‘₯. As the derivative of a number times a function is that number times the derivative of the function, all we have to do now is differentiate two to the π‘₯ with respect to π‘₯.

How do we differentiate the exponential function with base two, two to the π‘₯? Hopefully, we know about the number 𝑒, whose special property is that the derivative of 𝑒 to the π‘₯ with respect to π‘₯ is 𝑒 to the π‘₯. Whenever we differentiate an expression, where the variable we’re differentiating with respect to β€” in our case π‘₯ β€” appears in an exponent, then this is a fact we need to use. This means taking any exponential term that we want to differentiate and rewriting it so its base is 𝑒.

How do we do this with two to the π‘₯? Well, we can rewrite two as 𝑒 to the natural logarithm of two. And then using one of our exponent laws, we get that two to the π‘₯ is 𝑒 to the natural logarithm of two times π‘₯. We’ve made the base 𝑒. Now, how does that help? Well, we can apply the chain rule.

If we let 𝑧 equal the natural logarithm of two times π‘₯, then we need to find negative three times the derivative with respect to π‘₯ of 𝑒 to the 𝑧. Now applying the chain rule with 𝑓 equal to 𝑒 to the 𝑧, we get negative three times 𝑑 by 𝑑𝑧 of 𝑒 to the 𝑧 times 𝑑𝑧 by 𝑑π‘₯.

Let’s clear some space. Here, I’ve just copied the last line of working. What is 𝑑 by 𝑑𝑧 of 𝑒 to the 𝑧? Well, just like 𝑑 by 𝑑π‘₯ of 𝑒 to the π‘₯ is 𝑒 to the π‘₯ or 𝑑 by π‘‘π‘ž of 𝑒 to the π‘ž is 𝑒 to the π‘ž, 𝑑 by 𝑑 𝑧 of 𝑒 to the 𝑧 is 𝑒 to the 𝑧. And how about 𝑑𝑧 by 𝑑π‘₯? Well, 𝑧 is the natural logarithm of two times π‘₯. So 𝑑𝑧 by 𝑑π‘₯ is just the natural logarithm of two.

Are we done? Well, not quite, we have 𝑑𝑦 by 𝑑π‘₯ in terms of 𝑧 and would really rather prefer it to be in terms of π‘₯. We can substitute the natural logarithm of two times π‘₯ for 𝑧. And now, we have our answer written in terms of π‘₯. But we can do even better.

Earlier, we showed that we could rewrite two to the π‘₯ with a base of 𝑒 as 𝑒 to the natural logarithm of two times π‘₯. Now, we can do the reverse β€” rewriting 𝑒 to the natural logarithm of two times π‘₯ as two to the π‘₯. And so our final answer is negative three times two to the π‘₯ times natural logarithm of two.

In general, if you want to differentiate an expression involving two to the π‘₯ or three to the π‘₯ or even as you might see later π‘₯ to the π‘₯, you should first write that exponential with a base of 𝑒. This will probably involve using the natural logarithm function and some laws of exponents. Having written all the exponentials with a base of 𝑒, you can then differentiate using the chain rule.

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