Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Natural logarithms definition. The natural logarithm function L&X equals the integral of one to X of 1 divided by TDT where X is greater than 0. So the derivative of L&X by definition is really just the derivative of an integral. Since L&X equaled the integral, so the derivative of L&X is going to equal the derivative of the integral from 1 to X of 1 divided by TDT. So when we have the derivative of an integral, by the fundamental theorem of calculus, we know that we can put in the upper bound, so 1 / X times the derivative of X minus putting in the lower bound 1 / 1 times the derivative of the lower bound, so 1 / X. The derivative in terms of X of X is 1 -. 1 / 1 is 1 derivative of a constant zero, so the derivative of the natural log is just 1 / X. We're using the chain rule. If instead of the natural log of X, it was the natural log of a function, we'd have the derivative in terms of X of LNU equaling 1 / U times the derivative of U in terms of X. An example, the derivative of the natural log of X to the fourth minus six X ^2 - 3 = 1 / X to the 4th minus six X ^2 - 3 times the derivative of that X to the fourth minus six X ^2 - 3 using the chain rule. So we're going to now use the power rule. We're going to bring down the exponent. Take it to the one less power, SO4X cubed minus twelve X / X to the 4th minus six X ^2 -, 3. Remembering that the derivative of a constant is always 0 product rule, we're going to take the derivative of LNX equaling 1 / X. We're going to do a proof here. So if we took the derivative of lna of X using that chain rule, we'd have 1 divided by AX times the derivative of AX. The derivative in terms of X of AX, where a is just a constant is going to be a. So one over AX times a is going to cancel the 1 / X. So what's just happened is we have two different functions. We have the function LNX and the function of lane of A of X having the same derivative. Well, and anytime 2 functions have the same derivative, the original functions have to be equal with the exception of some constant. So we're going to have LNX plus that constant C equaling LNAX. And what our goal is to do is to find what that C would equal. Now this has to be true for every single value of X where X is greater than or equal to no X is greater than 0. So we're going to let X equal 1 because we know that the natural log of one is 0. So if we said let X1, we'd have natural log of 1 + C equaling the natural log of the quantity a * 1. In this case, natural log of one is 0, so we get C equaling LNA. So with this, we now have a property that says the natural log of X plus the natural log of A equals the natural log of AX. This is called the product rule. You've seen it in your algebra classes in the past. Whenever we have two things multiplied together that we're taking a logarithm of, we can rewrite it as the logarithm of each of them separately with addition. Remember that the natural log of 1 is really just log base E. Or we could think of it as E to sum power equaling one if we changed it into exponential form. Well, anything to the 0 powers one with the exception of 0 to 0, so M had to be 0, so the natural log of one is 0. We're going to look at a couple more properties. The next property we're going to look at is the difference rule, so the derivative of lane of X / a. Well, our property for the derivative of a natural log says 1 divided by whatever was in the parentheses. So 1 / X / a times the derivative of what was in the parentheses by that chain rule X / a. So 1 / X / a is really just it's reciprocal, or a / X the multiplicative reciprocal. The derivative of X / a is just going to be 1 / a because we could think of this as a coefficient of 1 / a * X. So then the A's will cancel. We get 1 / X if we take the derivative of just plain old natural log of X. Or we could think of it as X / 1. That's 1 / X by our earlier in this video. So now we have 2 functions that have the same derivatives, so we know that those two functions originally had to be exactly the same with some constant. So the natural log of X / a equals the natural log of X + C. This has to be true for every value greater than 0. So again we're going to let X equal 1 because we know the natural log of one is 0, So we get the natural log of 1 / a equaling natural log of one which is 0 + C or C is just the natural log of 1 / a. So the natural log of X / a equal the natural log of X plus the natural log of 1 / a 1 / a we can think of as A to the -1 power, and this -1 by the power rule can come down in front. Remember, 1 / a = A to the -1 and then by the power rule, we're going to bring that down in front. So the natural log of X / a equals the natural log of X minus the natural log of a. The power rule, the derivative of the natural log of X to some power is going to equal. We're going to take 1 divided by the X to the R times the derivative of X to the R by the chain rule, the derivative of this X to the R we're going to bring down the exponent and take it to the one less power. So we have that r * X to the r -, 1. Well, if we have X to the R on bottom and X to the r -, 1 in the numerator, we know that when we have the same basis and numerator and denominator, we actually subtract the exponents. So the exponent in the numerator was r -, 1 and we're subtracting R, The Rs will cancel. We get X to the -1 left, which is just really 1 / X. So this derivative simplifies into that constant r / X. And now if we take the derivative of RL and X, this R is just a constant. So the constant can come in and out of the derivative. So we have R times the derivative of L&X, which is just 1 / X. Two functions that have the same derivative have to vary by a constant. So lane X to the r = r, lane X plus some constant C Once again, we're going to let X equal 1. If X = 1, we get an L and one in here, which is 0. We also get one to some power, and we know that one to any powers 1, so that's also zero. So in this case, our constants just 0. So what this product rule says is that or the power rule? Sorry, the power rule says that when I have the natural log of X to some power, I can bring the power down in front and multiply it. So the graph of Y equal L&X when X is greater than 0, well, we know that the derivative of L&X is 1 / X and if X is greater than 0, we know this is always positive. Thus the original function is always increasing. If we take the derivative of 1 / X, we get -1 / X ^2. By the power rule, we have X to the -1 bring down the exponent to the one less power. It's always negative. Thus we know that the original function is going to be concave down the entire time. If we think about Y equaling LNX when X is one, we get y = 0. So we have an intercept at the .10. So we know that we're always increasing and we're always concave down. We also know that we're going to have an asymptote at X = 0, so the domain is going to be 0 to Infinity. The range is going to be negative Infinity to Infinity. If we think about limit as X goes to 0 from the right hand side of the natural log of X, that's negative Infinity which we see in our graph. If we think about the limit as X goes to Infinity, Infinity of natural log of X, that's getting bigger and bigger and bigger. Even though it's going gradually, it's still getting bigger. So Infinity if U is a differential function that is never zero, we have the integral of one over Udu which equals the natural log of the absolute value of U + C Remember we have to have that U being positive. So if we look at the integral of tangent XDX, we could think of tangent as sine over cosine and now we can do some U substitution. If we thought about U equaling cosine X, then the derivative of U would be negative sine XDX or negative DU as sine XDX. So right here's the sine XDX. I'm going to take it out and I'm going to substitute in negative DU instead of the cosine X. I'm going to put in U. So I have an integral of negative DU over U. I'm going to pull out the -1 in front. DU over U is really the same thing as 1 / U times DU. And we know that that's just the negative natural log of the absolute value of U + C. We always have that constant because we don't have a bounded integral. EU is cosine of X. So we're going to do a substitution here. So we're going to get negative lane at the absolute value cosine X + C. Now we could take this negative up on top by that power rule of logarithms that we developed a moment ago. If I took the negative on top, we know that cosine X to the -1 is 1 divided by cosine, while 1 divided by cosine is really the trig function of secant. So the integral of tangent XDX is equal to negative lane absolute value cosine X + C Or we could think of that as the natural log of the absolute value of secant X + C. That's one that you're going to want to remember. We're going to also do cotangent X. So if we have the integral of cotangent XDX, this time it's going to be cosine X over sine XDX. Let's U be the sine X. We're letting you be what's in the denominator, because if I have something divided by that denominator, hopefully we can use the natural logarithm integral. So DU is going to be cosine XDX. Doing our substitution, we get the integral of DU over U, which is just the natural log of the absolute value of U + C That U, we said by U substitution was just really sine X. So we have the natural log of the absolute value of sine X + C So this is one of the two that we could use. The other one comes from thinking about possibly taking sine X to the -1 to the -1 because powers to powers and multiply, that'd be the same thing as a +1 if I took this -1 down in front. And then sine X to the -1 is really just cosecant X. So a different way of writing the natural log of the absolute value sine X + C is the negative of the natural log of the absolute value of cosine X + C. So those are both equivalents. Here. The integral cotangent XDX equals the natural log absolute value sine X + C, which equals the negative of the natural log of the absolute value of cosine X + C. Let's look at some examples next. So if I wanted to find the derivative of X to the fourth over 4 lane X -, X to the 4 / 16, I'm going to have to do the product rule in this first part. So I'm going to take the derivative of the first times the second plus the derivative of the second times, the 1st. And then we're going to use the power rule for the second part where we bring down the exponent, take it to 1 less power. So X to the fourth over 4, we're going to bring down the four. We're going to take it to 1 less power times the second term plus the derivative of LNX, which is 1 / X times the first term minus bringing down the four and taking it to 1 less power. And then we're just going to simplify things up. So X ^3 LNX plus X ^3 / 4 -, X ^3 / 4, those last two terms are going to cancel. So we're just going to end up with a final answer of DYDXS X ^3 LNX. What if we have a composition function lane of lane X? So we're going to take the derivative of the whole thing. So the derivative of this lane is 1 divided by lane X. But then we need to take the derivative of this inside piece, so we're going to multiply it by the derivative of lane X. This is the chain rule. Remember, the derivative of lane X is 1 / X. So our final answer to this example would be DYDX is 1 / X lane X. What if we have a composition? Well, our rules that we developed a moment ago about natural logarithms is going to help us make this easier. If I have Y equal 1/2 natural log of 1 + X / 1 minus XI could actually instead of having division here and having to do the chain rule and the quotient rule both, I can rewrite that before I ever look at the derivative as 1/2 natural log of 1 + X -, 1/2 natural log of 1 -, X. So now when I find the derivative, I'm finding the derivative of each piece 1/2. The derivative of a natural log is 1 / 1 + X times the derivative of whatever was in the ( 1 + X. In this case -1 half natural log derivative is 1 divided by what was in the parenthesis times the derivative of what was in the parenthesis. When we have this simplified up, we get 1 / 2 * 1 + X + 1 / 2 * 1 - X. Getting a common denominator and simplifying, we end up with 1 / 1 + X * 1 - X. This next one we're going to find the integral this time of eight XDX over four X ^2 - 5. We're going to do AU sub. We're going to let U equal four X ^2 -, 5 S Our DU is 8 XDX. When we substitute, we get the integral of DU over U, which we know is the natural log of the absolute value of U + C And then we're just going to do the substitution back. So natural log of absolute value four X ^2 -, 5 + C. Thank you and have a wonderful day.