,, . I can cook up some really nasty functions, give them to you. And then ask
you to differentiate them. For instance, I could ask you to differentiate this
function. 1 plus x squared to the 5th power, 1 plus x cubed to the 8th power,
all divided by 1 plus x to the fourth, this to the seventh power. In principle,
there's nothing stopping you from plowing ahead and computing the derivative. You
can totally differentiate this function. Right? What's the derivative of this
function? Well, this function's a quotient so he needs the quotient rule. The
denominator of the quotient rule is the original denominator squared. So it's
going to be the original denominator now to the 14th power And the quotient rule,
the numerator starts off with the derivative of the original numerator. Now,
the original numerator is a product. So I'll be able to do this derivative by
using the, product rule and chain rule. So it's the derivative of the numerator imes
the denominator. And it keeps going, right? Then I gotta subtract the,
derivative of the denominator times the numerator. But, look, you can do this
derivative just by careful application of the quotient rule, the product rule, the
power rule, and the chain rule. There is one thing stopping you, your sense of
human decency. It's just an awful calculation. Nobody would want to do that.
So instead, I propose a trick. But maybe it's not a trick, because it's a trick
that fits into a general theme. It's logarithms. Logarithms turn exponentiation
into multiplication, and multiplication into addition. Let's see how this helps
us. So here we go. Instead of calling this function f of x. I'm jut going to call it
y, because I'm getting ready to do a sort of implicit differentiation. I'm going to
first apply log to both sides of this. And I'll get log y. And what's log of the
other side? Well it 's log of a quotient, which is a difference of logs, and logs of
things to powers, which is that power times log of the base. So this works out
to 5 times log of 1 plus x squared plus This log turns multiplication into
addition. 8 times log of 1 plus X cubed and this quotient becomes a difference,
so, minus 7. The 7 in the exponent, log 1 plus x. Now, we differentiate. All right,
so differentiating now, what's the derivative of log y? Remember, y is
secretly a function of x, so I differentiate log y. It's the derivative
of the outside, which is 1 over y, times the derivative of the inside, which, I'll
write dy dx. This is really an example if you like the implicit differentiation.
Alright, now I differentiate the other side, 5, I just multiply it by 5 the
derivative of log is 1 over, so 1 over the inside function and 1 plus x squared times
the derivative of the inside function which is 2x. The derivative of 1 plus x
squared is 2x. All right, plus 8, and its derivative log is 1 over at the inside
function, 1 plus x cubed, times the derivative of this inside function, which
is 3 times x squared, minus 7 over, the derivative of log is one over at the
inside function, one plus x to the fourth, and the derivative of one plus x to the
fourth is four x cubed. We're almost there. So now, I just multiply both sides
by y. And I get that the derivative is this thing calculated x y, y is this
quantity. I can write this a little bit more nicely alright here's this 5 times 2x
is 10x, 8 times 3 is 24, 7 times 4 is 28, and then I multiply by y. So I found the
derivative, here it is. In general, this trick logarithmic differentiation as it's
called, works fantastically well for functions like these. Rational functions
that involve a lot of high powers.