This is the proof I showed earlier that the derivative of the function ln x is 1/x. We start with the equation
Differentiating both sides with respect to x gives
or
Two things are interesting and useful about the natural log function. The first is that it has a relatively simple derivative. The second is that logarithms can turn products and quotients into sums and differences and that it can simplify expressions containing powers.
This combination of characteristics leads to a useful shortcut trick for computing derivatives called logarithmic differentiation. Here is an example of how this works. Consider the function
Computing the derivative of this monster by conventional means is a bit of a mess.
Logarithmic differentiation starts by taking a natural log on both side of the original equation and using the properties of the natural log to simplify wherever possible.
Differentiating both sides with respect to x gives
Substituting for f(x) and solving for 
gives
Suppose we know the value of a function and its derivative at some point. This is enough information to be able to compute a tangent line to the curve at that point. There are two ways to write the equation of a line.
y = m x + b
For the present application the second of these, the point slope formula, is more useful. The slope is given by the derivative at the point we are interested in.
The tangent line to a curve provides a useful approximation technique. Because the equation of the tangent line is so simple, it is relatively easy to do computations with it. Further, for points close to the original x0, the tangent line approximates the original curve reasonably well.
As an example, let us compute sin x for x near some known value. We know that at 
the function sin x has value 
. Can we use that knowledge to compute an approximation for sin
. The equation of the tangent line is
or equivalently
If we substitute in 
for x, we get the corresponding y value for the tangent line.
You can compare that with the exact value of the original sine function
to see that the approximation is quite good.
The approximation gets worse and worse as we move away from the point at which we drew the tangent line, as this graph of the original function and the tangent line shows.
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We have just seen that the most convenient way to construct the equation of the tangent line to a curve y = f(x) at the point (x0,y0) is to use the point-slope formula for a line.
For actually calculating things, you may find it more convenient to write this equation in explicit form.
This formula is called the linear approximation to f(x) at x0.
This time, let us estimate 
. 61.5 is close to 64, and we know that 
. Let us compute the linear approximation to the curve 
at the point (64,8) and use that to compute an estimate for 
. For this problem, 
, so 
. The point 
is 
, so the linear approximation is
The exact value of 
is
Back before we had computers and calculators to calculate things for us, an awful lot of work in mathematics involved calculating and estimating. One very important tool for computing estimates was known as interpolation. Over the centuries, mathematicians had developed many useful numerical methods for computing quantities to a high degree of accuracy. Often, the results of these calculations were published in tables bound in reference manuals.
For example, if you wanted to compute a square root, you would consult a table of square roots. Consider the 
example from above. A typical table of square roots would contain some entries like the following.
| x |  |
|---|---|
| 60 | 7.745967 |
| 62 | 7.874008 |
| 64 | 8.000000 |
Knowing that 61.5 is between 60 and 62 we know that its square root is somewhere between 7.75 and 7.87. To make the best estimate possible, we can compute the equation of the line connecting the two points (60 , 7.745967) and 
and then find the y value on that line that corresponds to x = 61.5. The slope of the line connecting the two points is
Let us call the point that we are trying to estimate 
. The slope of the line connecting 
to 
is the same as the line connecting 
to 
, so we have
Again, this is a fairly decent approximation.
Section 3.8: 8, 9, 38, 39, 47
Section 3.11: 6, 7, 32, 33