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,, . I can cook up some really nasty
functions, give them to you. And then ask

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you to differentiate them. For instance, I
could ask you to differentiate this

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function. 1 plus x squared to the 5th
power, 1 plus x cubed to the 8th power,

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all divided by 1 plus x to the fourth,
this to the seventh power. In principle,

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there's nothing stopping you from plowing
ahead and computing the derivative. You

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can totally differentiate this function.
Right? What's the derivative of this

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function? Well, this function's a quotient
so he needs the quotient rule. The

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denominator of the quotient rule is the
original denominator squared. So it's

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going to be the original denominator now
to the 14th power And the quotient rule,

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the numerator starts off with the
derivative of the original numerator. Now,

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the original numerator is a product. So
I'll be able to do this derivative by

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using the, product rule and chain rule. So
it's the derivative of the numerator imes

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the denominator. And it keeps going,
right? Then I gotta subtract the,

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derivative of the denominator times the
numerator. But, look, you can do this

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derivative just by careful application of
the quotient rule, the product rule, the

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power rule, and the chain rule. There is
one thing stopping you, your sense of

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human decency. It's just an awful
calculation. Nobody would want to do that.

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So instead, I propose a trick. But maybe
it's not a trick, because it's a trick

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that fits into a general theme. It's
logarithms. Logarithms turn exponentiation

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into multiplication, and multiplication
into addition. Let's see how this helps

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us. So here we go. Instead of calling this
function f of x. I'm jut going to call it

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y, because I'm getting ready to do a sort
of implicit differentiation. I'm going to

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first apply log to both sides of this. And
I'll get log y. And what's log of the

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other side? Well it 's log of a quotient,
which is a difference of logs, and logs of

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things to powers, which is that power
times log of the base. So this works out

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to 5 times log of 1 plus x squared plus
This log turns multiplication into

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addition. 8 times log of 1 plus X cubed
and this quotient becomes a difference,

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so, minus 7. The 7 in the exponent, log 1
plus x. Now, we differentiate. All right,

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so differentiating now, what's the
derivative of log y? Remember, y is

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secretly a function of x, so I
differentiate log y. It's the derivative

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of the outside, which is 1 over y, times
the derivative of the inside, which, I'll

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write dy dx. This is really an example if
you like the implicit differentiation.

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Alright, now I differentiate the other
side, 5, I just multiply it by 5 the

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derivative of log is 1 over, so 1 over the
inside function and 1 plus x squared times

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the derivative of the inside function
which is 2x. The derivative of 1 plus x

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squared is 2x. All right, plus 8, and its
derivative log is 1 over at the inside

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function, 1 plus x cubed, times the
derivative of this inside function, which

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is 3 times x squared, minus 7 over, the
derivative of log is one over at the

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inside function, one plus x to the fourth,
and the derivative of one plus x to the

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fourth is four x cubed. We're almost
there. So now, I just multiply both sides

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by y. And I get that the derivative is
this thing calculated x y, y is this

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quantity. I can write this a little bit
more nicely alright here's this 5 times 2x

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is 10x, 8 times 3 is 24, 7 times 4 is 28,
and then I multiply by y. So I found the

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derivative, here it is. In general, this
trick logarithmic differentiation as it's

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called, works fantastically well for
functions like these. Rational functions

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that involve a lot of high powers.