So in section 3.9, we're going to be discussing the derivatives of logarithmic and exponential functions. So as a reminder, in blue is the graph of an exponential function, and in red we have the graph of a logarithmic function. So these are the two basic graphs.
Again, remember that logs and exponentials are inverses of each other. So you can see relative to the two graphs here, what we have is essentially just an exchange. of the x and the y coordinates.
So the derivative of an exponential function first, now we've already talked about the derivative of log, excuse me, we've already talked about the derivative of e to the x, but now we're gonna talk about the derivative of b to the x, where b is any base. So again, we keep b larger than zero. We talked about this earlier in chapter.
one, we want b larger than zero so that we avoid complex numbers or imaginary numbers coming out of our exponential function. So the derivative of b to the x, again b is any number now, the derivative is going to be b to the x times the natural log of b. So we've already seen this rule applied for the case that the base is equal to e. Again, e to the x, the derivative is e to the x.
Applying this rule, we would write e to the x natural log e, but the natural log of e is just 1, so we just have e to the x. But now we have a rule for the general base. Any base b to the x power, the derivative is going to be b to the x times the natural log of b. Now we can't forget the chain rule, so this is essentially just a template.
So this is the template for the base derivative for b to the x, but the chain rule will always apply. So let's say that I have b to the g of x, where g of x is just some function. Well, we're going to have that base template, so we'll have b to the g of x times the natural log of b, but then we have to apply the chain rule and multiply by the derivative of g. So the chain rule never goes away and so this derivative rule here is essentially just a template. Whenever the chain rule applies we have to apply the chain rule.
So a very brief proof of how we end up with this derivative rule the derivative of b to the x being b to the x natural log b. So again we can form the limit definition of the derivative so we have f of x plus h minus f of x divided by h so that's b to the x plus h minus b to the x. Divide by h. So b to the x plus h is the same as b to the x times b to the h.
And now we can factor out a b to the x. As h goes to 0, well, b to the x has no relation to this variable h. So this b to the x term is behaving as a constant. And so we have now b to the x times this limit. The limit as h goes to 0 of b to the h minus 1 over h.
It can be shown that that limit is equal to the natural log of b, and we will get our result b to the x times the natural log of b. Again, as a reminder, you want to be careful to not mix up your derivative rules. So now we have this new derivative rule for b to the x.
So again, this is not the power rule. Our derivative is not x times b to the x minus 1. So we've got two situations. We've got a number to a variable. That's this exponential rule that we have now. We also have the case of a variable to a number.
That's the power rule. So be careful to keep your derivative rules straight and to not mix them together when it's not appropriate to. So now we'll introduce the derivative of a logarithmic function. So let y equal log base b of x. Again, b larger than 0. So what we have is that the derivative of log base b of x is equal to 1 over x times the natural log of b.
And we have the case that b equals e. Well, if we have b equals e, that's the natural log function. The natural log of e is, again, 1. So for the natural log of x, our derivative will simply just be 1 over x.
Again, these are just templates. If there's something more going on inside of these logarithmic functions, we have to apply the chain rule. So if I had, let's say, log base b of g of x, So again I apply the general template. The general template says 1 over the inside times the natural log of b. So I have 1 over g of x times the natural log of b.
And we must finish it with the chain rule times g prime of x. So the chain rule will never go away. It is always present and it may always need to be applied depending on the function that we have.
So let's go briefly through this rule that we have for the logarithmic function so we can show this rule through implicit differentiation. So let's go through that together. So we'll start with our logarithmic function.
So let y equal log base b of x. So what we can say is that this is the same. as b to the y equals x. So this is now an equation that's no longer solved for y. So let's apply implicit differentiation.
We know the derivative of b to the y is b to the y times the natural log of b. But I took the derivative of a y term. That y term is dependent on x. So we now attack on dy dx.
And on the right-hand side, the derivative is just going to be equal to 1. So we'll just solve for dy dx now. So dy dx, dividing by this expression, 1 over b to the y, natural log b. But what do we have about b to the y? b to the y is equal to x. So we can now say that the derivative dy dx is going to be 1 over x natural log b.
So let's go through some examples and apply the new rules that we have now. Our function here is y equals x squared plus 2 to the x plus log base 2 of x minus the natural log of x. So I want to find dy dx.
Of course, that's the derivative, so dy dx is simply going to be term by term. The derivative of x squared is, of course, 2x. Now for the derivative of 2 to the x, I'm going to have 2 to the x natural log 2. So that's our b to the x. The derivative of b to the x is b to the x ln b. In this case, we have 2. Now we have the derivative of log base 2 of x.
The derivative of log base 2 of x is 1 over x, natural log 2. And then minus the derivative of the natural log of x, which is 1 over x. So all one line here, this is the derivative of this function. Again, the actual expression here would be 1 over x natural log e if we wanted to, but again, natural log e is equal to 1. The next function here is y equals 10 to the x cubed minus 3x power.
So we'll find the derivative, y prime. So notice that we're going to have to apply the chain rule. I've got this function x cubed minus 3x shoved into the function 10 to the x.
So we have a very clear composition of functions that's going to require the chain rule. So we're going to have our template. So our template says we have b to the x. times the natural log of b, so times the natural log of 10. But now that I have this function x cubed minus 3x shoved inside, I've got to multiply by its derivative, which is, of course, 3x squared minus 3. So this expression here, y prime, is going to be the derivative of this function. Next, we have the function y equals 2 to the x divided by 3 to the x plus 4 to the x.
So notice I have two different, I've got... three different exponential functions, 2 to the x, 3 to the x, and 4 to the x, and we have a division going on, so that's going to require the quotient rule. So the quotient rule says we'll take the derivative of the top, that's going to be 2 to the x natural log 2, times the bottom, so we'll have 3 to the x plus 4 to the x, and then minus the top times the derivative of the bottom, well the derivative of the bottom. It's going to be 3 to the x natural log 3 plus 4 to the x natural log 4. And of course, quotient rule, that all gets divided by 3 to the x plus 4 to the x quantity squared. Here our function reads y equals the natural log of kx, where k is any real number which is not equal to 0. So we'd like to find the derivative dy dx.
So the derivative dy dx is going to be simply 1 over kx. So again, for the natural log, we have base e. So we can sort of forget that l and e term at the bottom since it's equal to 1. So I have 1 over kx. Now I have to multiply by the derivative of the inside of this natural log function, the derivative of.
The function kx would just be k. So notice these k's will cancel. So our derivative of this function, natural log kx, is simply 1 over x.
So it seems that the k is actually irrelevant to our derivative because it's always going to cancel. And what you might also notice here is that properties of logarithms, natural log kx, is natural log k plus natural log x. And so then the derivative of natural log k plus natural log x, well natural log k is just a number so its derivative is zero.
And so we're left with 1 over x being the derivative of the natural log of kx through this direction as well. The function we have here is the function natural log of x e to the x minus e to the x and we want to find the derivative. So the derivative for the natural log will be 1 over the inside.
Now notice that the inside itself is some non-trivial function. So we want to multiply by the derivative of the inside. The first part here is going to require the product rule. So the derivative will be e to the x plus x e to the x plus x. the x.
That's the product rule component. Now the derivative of minus e to the x will be minus e to the x. So notice we get a nice cancellation.
So our derivative is going to be simply the expression x e to the x over x e to the x minus e to the x. And you might notice here that we could also make a cancellation. Notice that these e to the x's are going to cancel.
And so our derivative could also be expressed as simply x over x minus 1. So that's interesting, the e to the x can be dropped out completely from this derivative. On occasion, we may be able to use properties of logarithms to simplify a function before finding the derivative. So remember the two properties of logarithms as it pertains to products and quotients inside of a logarithm. Remember that the product can be split into a sum, and then a quotient can be split into a difference.
So if I have this function here, the function y equals the natural log of x, excuse me, the natural log of the sine of x times the cosine of x, Well, this is the same as the natural log of the sine plus the natural log of the cosine. So I don't have to apply the property here, but the property will certainly make things a little bit easier. I'll be able to avoid the product rule when I apply the chain rule. So now our derivative would just be y prime.
The derivative of natural log sine is going to be 1 over sine. So 1 over the inside now times the derivative of the inside, which is of course cosine. Now the second term will have 1 over the inside times the derivative of the inside, which is going to be negative sine. So perhaps you'll notice here that sine over, excuse me, cosine over sine, that's going to be cotangent. So our derivative can be written as cotangent.
Now minus sine over cosine, that's going to be minus tangent. So our derivative is very simply the cotangent minus the tangent. And this was a bit simpler than having to apply the product rule when we go through the application of the chain rule. So on occasion, we...
may be able to apply properties of logarithms to simplify our work a little bit. Also let's look at this function, the function f of x equals the natural log of x squared plus one over x squared minus one. Again, we've got a quotient inside of a logarithm, so this is the same as the natural log of x squared plus one minus the natural log of x squared minus one. So this is nice because I'm not going to have to apply the quotient rule when I apply the chain rule. So the derivative now is going to be simply the first term, 1 over the inside, times the derivative of the inside, which is 2x, and then minus 1 over the inside times the derivative of the inside, also 2x.
So we can write this derivative as simply... 2x over x squared plus 1 minus 2x over x squared minus 1. Alright, and certainly I believe you would agree that's going to be much simpler than having to apply the quotient rule. So when possible, feel free to apply properties of logarithms to simplify your work a little bit. In this example, For example, I'd like to find h prime of the natural log of 2 given that h of x is the natural log of e to the 2x plus 1. So the derivative will just be 1 over the inside and then times the derivative of the inside which is going to be 2e to the 2x. So our derivative is simply 2e to the 2x.
over e to the 2x plus 1. So we want to find h prime of the natural log of 2. So that's going to be 2e to the 2 times natural log 2, all divided by e to the 2 natural log 2 plus 1. So now a couple of things we can do here. We need to simplify this. Remember that the property of logarithms that allows us to bring a power out as a coefficient, we can reverse that and put the power back. or put the coefficient back as a power.
So this would be 2e to the natural log of 2 squared, which would be 4. And then I'll do the same thing here, e natural log 4 plus 1. So what do we have left over now? Well, the e to the natural log 4 expression is just going to be equal to 4. So on the top, I have 2 times 4 divided by 4 plus 1. And that's going to be 8 divided by 5. So in this expression I took this coefficient and moved it back as a power. 2 squared is of course 4. E natural log 4 is 4. And so we have our final result here simplified to the very simple fraction 8 over 5. So one final function before moving forward. Here is the function y equals log base 4 of e to the 2x plus 3 to the x squared.
So the chain rule abounds in this example. I'm going to have to apply the chain rule. for the function log base 4 with this function inside. And as I apply the chain rule, this piece requires the chain rule, and so does this piece.
So our derivative is going to be y prime. So for log base 4, our template is 1 over the inside. And then natural log 4 on the bottom.
and then times the derivative of the inside. So the derivative of e to the 2x is going to be 2e to the 2x by the chain rule. And then the derivative of 3 to the x squared is going to be 3 to the x squared natural log 3. And then we have to hop in and finish by multiplying by the derivative of the inside. And of course the derivative of x squared is 2x. So I'll express our derivative this way.
What we have is 2e to the 2x and then plus 3 to the x squared natural log 3 times 2x all divided by e to the 2x plus 3 to the x squared times the natural log of 4 on the bottom. So to finish out this section we introduced the technique of logarithmic differentiation. So let's say that we were posed with a function of this form. y equals x to the x power.
So notice that I have both a variable in the base and a variable in the exponent. So this function is sort of like a hybrid between an exponential function and a power function. A power function is something like x squared. An exponential function is something like 2 to the x. So this is like a mixture of those two.
I have y equals x to the x power, so a variable raised to. another variable. So how do we find the derivative in this case? Well first of all, this is not going to be an application of the power rule.
So the derivative is not going to be bring the power of x down, take one away, because this is not a power function. A power function says we have x to a number. This is also not going to be a logarithmic function, excuse me, an exponential function, because an exponential function says we have a number to a variable. And so I cannot say that the derivative here would just be x to the x, l, and x, doing our b to the x, l, and b pattern. So neither of these cases apply to this function.
So for these types of functions where we have a variable to a variable, we're going to employ the technique of what we call logarithmic differentiation. And the way that we are going to go forward is we're going to To make use of this fact, the natural log of x to a power is the same as that power being written as a coefficient. So this step right here is going to be critical, or applying this property is going to be critical in moving forward in this technique of logarithmic differentiation.
So let's see how we find the derivative of this function y equals x to the x power. So what we're going to do here... is we're going to take the natural log of both sides. So we'll take the natural log of both sides of this function.
We'll have the natural log of y is equal to the natural log of x to the x. And so this is the same thing now as the natural log of y. Again, what we'll do is we'll apply that property.
This x can come out to the front. So we have natural log y is equal to x natural log x. Now, What we've done here is we're no longer solved for y. So this may hint that we're going to go forward using implicit differentiation. But now I've rewritten this right-hand side into a form that looks a lot more familiar.
This is now just simply a product. And so I can proceed now by implicit differentiation. So we'll take the derivative of both sides. So what we're going to have is the derivative of natural log y.
is going to be 1 over y, but y is a function of x as we're seeing here, so we have to apply the chain rule and multiply by dy dx. So we have 1 over y dy dx on the left-hand side. Now on the right-hand side, we're going to apply the product rule.
So we've got our two functions, x and the natural log of x. So f prime g, we'll write 1 times the natural log of x, and then plus f g. prime the derivative of the natural log of x is going to be 1 over x.
So we'll rewrite this one more time before we start to solve for dy dx. So this is going to be 1 over y dy dx equals the natural log of x plus 1. Right? x over x is of course just going to be equal to 1. So what we want to do now is we want to isolate dy dx.
So what we're going to do is we're going to multiply both sides by y. And so what we have is that dy dx is going to be equal to y times the natural log of x plus 1. Now how do we want to finish this? So on the left-hand side, these y over y's are going to cancel, even if we just dy over dx on the left-hand side.
How do we want to finish this? Well, we would like our derivative to be completely in terms of x. So I have here...
that y equals x to the x. So this is the same y. So we can now just replace this with x to the x.
So we now have our result. The derivative of x to the x is going to be x to the x times the natural log of x plus 1. Right and we achieve that result by logarithmic differentiation. So let's walk through that one more time.
So we have this function y equals x to the x power. So I identify the variable in both the base and the exponent. I'll take the natural log of both sides and apply this property and bring this power of x down as a coefficient.
So we've finished this by applying effectively now we're just doing implicit differentiation. So now the derivative of l and y is 1 over y dy dx. The derivative of the right hand side by the product rule becomes L and x plus 1. We want to isolate dy dx, so we multiply across by y. And then we finish it by replacing y with its definition, which is x to the x.
And so here's our final derivative. The derivative of x to the x is x to the x times the natural log of x plus 1. Again, we achieve that through this technique of logarithmic differentiation. So notice here that this result. is very different than the two results we had here. So this goes to immediately show that these two statements here are false.
So whenever you can identify a variable in both the base and the exponent, it's going to require this technique of logarithmic differentiation. So let's go through this one more time. This function now is x to the sine of pi x. So again notice that I have Variable in the base, variable in the exponent. So we'll have the natural log of y, taking the natural log of both sides, is the natural log of x to the sine of pi x, applying the properties of logarithms, we can take the sine of pi x.
and move it to the front. So this becomes sine pi x times the natural log of x. So that's the statement we'll work from now.
So again we're just going to proceed by implicit differentiation. So by implicit differentiation the derivative of the left hand side is 1 over y dy dx. Now the derivative of the right hand side we're going to again apply the product rule.
So the derivative of sine of pi x is going to be cosine of pi x times pi and then times the natural log of x. So there's my f prime g and then we'll have f g prime. So let's rewrite this one more step before solving for dy dx.
So I'm just going to move some things around so they flow together a little better. So this is pi cosine pi x times the natural log of x. plus the sine of pi x divided by x. So again we want to multiply both sides by y.
So on the left hand side we now have isolated dy dx and this is going to be equal to y times our expression pi cosine pi x natural log x. plus the sine of pi x over pi, excuse me, over x. And so we'll finish this by just replacing y with its definition. So again, y is x to the sine of pi x. So we're just going to make that replacement.
So everything is back in terms of x. So y is x to the sine of pi x. And so this expression here at the end is our derivative.
Again, through this technique of implicit differentiation. Excuse me, this technique of logarithmic differentiation, which you're seeing is essentially implicit differentiation after one manipulation using properties of logarithms.