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# Derivative of sin(ln(x²))

We apply the chain rule twice to differentiate sin(ln(x²)), breaking down this composite function into simpler elements. This double application allows us to handle the complexity of the function, turning it into manageable parts. Together, we'll tackle this mathematical task with precision and clarity. Created by Sal Khan.

## Want to join the conversation?

- At2:25, could you also write g'(h(x)) as (x^2)^-1, which would by the power rule would turn out to be -(x^2)^-2 . Am I wrong?(14 votes)
- Close, but wrong.

1). You're solving a different problem. You're trying to get the derivative of 1/(x^2), which as you stated can be rewritten as (x^2)^-1. But Sal was getting the derivative of ln(x^2).... but to continue on with (x^2)^-1, and why the derivative of that isn't -(x^2)^-2...

g(x) = x^-1, h(x)=x^2 derivative of g(h(x)) = g'(h(x))*h'(x) by the Chain Rule

you forgot to multiply by h'(x).

So you got half of that correct, g'(x) = -1(x)^-2, by the Power Rule.

and g'(h(x)) = -1(x^2)^-2

h'(x) = 2x, also by the power rule.

By the Chain Rule the derivative is of (x^2)^-1 is -1((x^2)^-2)*2x, you forgot the 2x.

2). The inner composite sal was solving is ln(x^2), not (x^2)^-1.

g(x) = ln(x), g'(x) = 1/x

h(x) = x^2, by the power rule, h'(x) = 2x

g'(h(x)) = 1/x^2, or (x^2)^-1

g'(h(x))*h'(x) = (1/(x^2))*2x = 2/x

And that's just the derivative of the inner composite.

Hope that helped.(32 votes)

- Can the chain rule be applied backward to anti derivatives?(17 votes)
- Indeed it can. This technique is often referred to as "u-substitution". For example, we know that the derivative of sin^2(x) is 2*sinx*cosx. Now, if I want to take the antiderivative of that, we have

∫ 2*sinx*cosx dx I can make the substitution u=sinx and du/dx=cosx or, equivalently, du=cosx*dx

∫2*u*du = u^2 + c = (sinx)^2 + c(18 votes)

- what is the composite of a function?(6 votes)
- Hello!

Composition of functions basically means that you**replace the placeholder x in f(x) with a function**.`f(x) = x^2`

, could be rewritten as:`f("our number") = "our number"^2`

.

Similarly, we could replace the placeholder`x`

with a "proper" function. Let me define:`g(x) = 2x + 2`

.

If I were to replace the placeholder`x`

in`f(x)`

with`g(x)`

, I would get:`f(g(x)) = (2x + 2)^2`

.

This is basically the same principle as:`f(5) = 5^2`

.

We replaced our placeholder x with 5, and that number becomes squared, since`f(x) = x^2`

.

Composition of functions is when you replace the placeholder in`f(x)`

with another function`g(x)`

.

Finally,`f(g(x))`

can be written as`f o g(x)`

, which can be interpreted as**applying the function g to the function f**.

Hope I could help!

// Kristian(21 votes)

- Can anyone who is proficient in this subject clarify this problem for me? Say, we have
`d/dx f[g(h(x))]`

as the example in this video. Is it equivalent to`d/d[g(h(x))] f[g(h(x))] * d/d(h(x)) g(h(x)) * d/dx h(x)`

?(6 votes)- Not quite. It should be:
`d/dx(f(g(h(x))) = f'(g(h(x))) * g'(h(x)) * h'(x)`

, where I'm using the "prime" notation for the derivative instead of d/dx. One way to think about it is that the Chain Rule "chains" together the derivatives of whatever functions are being composed. So if you have the composition of three functions f, g, and h, you should expect your derivative for`f(g(h(x)))`

to be the product of three derivatives (as shown above). Hope that helps!(8 votes)

- In the exercise that follows, how is that

2•csc(x)•sec(x) + 2x•-csc(x)•cot(x)•sec(x) + 2x•csc(x)•sec(x)•tan(x)

is becoming

2•csc(x)•sec(x) - 2x•csc^2(x) + 2x•sec^2(x)

?

I mean, what happened to cot(x) and tan(x)?(4 votes)- I found out what happened. If anyone is lost, it comes from:

sin(x) = cos(x)tan(x)

cos(x) = sin(x)cot(x)

tan(x) = sec(x)sin(x)**csc**(x) =**cot**(x)**sec**(x) <--------**sec**(x) =**tan**(x)**csc**(x) <--------

cot(x) = csc(x)cos(x)(4 votes)

- Is there a recursive formula that can give us an arbitrary or even infinite number of functions to be chained together to be derived? In other words, is there a chain rule of _k_ or infinite composition?(4 votes)
- Sure, and this is how a computer algebra system handles it. Let's start a little simpler: Say you've got a simple addition, a + b + c + d. And if you only knew how to multiple two numbers, you could approach it as a + (b + (c + d))). That's called "reduction," where you take many operations and break them into parts.

Here's where it gets tricky: f(g(x)) can be viewed as f • g, and f(g(h(x))) is like f • (g • h), where the dot • is called "function application". We take away the x’s because we're dealing with the functions themselves as values.

That makes it a bit more obvious that the chain rule is an operation on that application, it's just saying, (f • g)’ = (f’ • g) ⨉ g’ so we're just moving some symbols around.

And then if I have (f • (g • (h • i)))’, you can see how you'd just churn through it from the outside in:

q = (g • (h • i))

(f • (g • (h • i)))’ = f’ • (g • (h • i))) ⨉ (g • (h • i))’

(g • (h • i))’ = g’ • (h • i) ⨉ (h • i)’

(h • i)’ = h’ • i ⨉ i’

And then substitute back in:

(g • (h • i))’ = g’ • (h • i) ⨉ (h’ • i ⨉ i’)

(f • (g • (h • i)))’ = f’ • (g • (h • i))) ⨉ g’ • (h • i) ⨉ h’ • i ⨉ i’

Anyway, if you had an arbitrary array of functions k[1], k[2], k[3]..., it looks like your rule would be:

(k[1] • k[2] • k[3] … k[n])’ = ∏ k[i]’ • k[i-1] • k[i-2] … k[1] where i goes from 2 to n

Or in more traditional notation:

d/dx (k[1](k[2](k[3](...k[n](x)…)))) = ∏ k[i]’(k[i-1](k[i-2](…k[1](x)…))) where i goes from 2 to n(4 votes)

- i have a doubt.. can we take the derivative of ln (2x) with chain rule? or its just 1/2x.. how do you build the composite function??(2 votes)
- the derivative is
`1/(2x) * 2`

which simplifies to`1/x`

.`1/(2x)`

is the outer derivative, and here it is important to leave the inner function`2x`

unchanged!`2`

is the derivative of the inner function`2x`

.(5 votes)

- At2:21Sal says that the derivative of lnx is 1/x. How does that work? And would it be similar for finding the derivative of a log?(3 votes)
- ln(x) is the natural logarithmic function. The proof of (ln(x))'=1/x is at the end of the content list.(3 votes)

- Does the same pattern show up for longer chains? So if you had to take the derivative of f(g(h(k(x))) would it be f'(g(h(k(x))) * g'(h(k(x))) * h'(k(x)) * k'(x)?(3 votes)
- Can't we just use 2ln(x), instead of ln(x^2), based on the properties of logarithms, and save ourselves an extra chain rule step?(2 votes)
- Of course, you get the same derivative value for both. It's perfectly alright as a method but not from a graphical standpoint and shouldn't the derivative value be supported by the graph? You have to remember that even though the value of ln(x^2) is 2ln(x), as functions, the two are different. The derivative of ln(x^2) at -1 is -2. This makes sense even on the graph. But does it make sense for 2ln(x) to have a slope of -2 at negative one. Can 2ln(x) even attain that value?

Since the functions ln(x^2) and 2ln(x) have different domains, they are essentially different functions.(2 votes)

## Video transcript

So now we're going
to attempt to take the derivative of the sine of
the natural log of x squared. So now we have a function that's
the composite of a function, that's a composite
of another function. So one way you
could think of it, if you set f of x as being equal
to sine of x, and g of x being the natural log of x, and
h of x equaling x squared. Then this thing right over
here is the exact same thing as trying to take the
derivative with respect to x of f of g of h of x. And what I want to
do is kind of think about how I would
do it in my head, without having to write all
the chain rule notation. So the way I would
think about this, if I were doing this in
my head, is the derivative of this outer function
of f, with respect to the level of composition
directly below it. So the derivative of
sine of x is cosine of x. But instead of it
being a cosine of x, it's going to be cosine of
whatever was inside of here. So it's going to be
cosine of natural log-- let me write that
in that same color-- cosine of natural
log of x squared. I'm going to do x that
same yellow color. And so you could really view
this, this part what I just read over here, as f prime,
this is f prime of g of h of x. This is f prime of g of h of x. If you want to keep
track of things. So I just took the derivative
of the outer with respect to whatever was inside of it. And now I have to
take the derivative of the inside with respect to x. But now we have another
composite function. So we're going to
multiply this times, we're going to do the
chain rule again. We're going to take the
derivative of ln with respect to x squared. So the derivative
of ln of x is 1/x. But now we're going
to have 1 over not x, but 1 over x squared. So to be clear, this part right
over here is g prime of not x. If it was g prime of x,
this would be 1 over x. But instead of an x, we
have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we
can take the derivative of our inner function. Let me write it. So we could write this
is g prime of h of x. And finally, we
just have to take the derivative of our innermost
function with respect to x. So the derivative of
2x with respect to-- or the derivative of x squared
with respect to x is 2x. So times h prime of x. Let me make everything clear. So what we have right
over here in purple, this, this, and this
are the same things. One expressed concretely,
one expressed abstractly. This, this, and this
are the same thing, expressed concretely
and abstractly. And then finally, this and
this are the same thing, expressed concretely
and abstractly. But then we're done. All we have to do to be done
is to just simplify this. So if we just change the order
in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this x over x,
2x over x squared is the same thing as 2 over x. And we're multiplying it
times all of this business. So we're left with 2 over x. And this goes away. 2 over x times the cosine of
the natural log of x squared. So it seemed like a very
daunting derivative. But we just say, OK, what's the
derivative of sine of something with respect to that something? Well, that's cosine
of that something. And then we go in
one layer, what's the derivative of
that something? Well, in that something we
have another composition. So the derivative of ln of x,
or ln of something with respect to another something,
well that's going to be 1 over
the something. So we had gotten a 1
over x squared here, that squared got canceled out. And then finally, the derivative
of this innermost function, it's kind of like
peeling an onion. The derivative of
this inner function with respect to x,
which was just 2x. Which we got right over here. This was 1 over x squared. This was 2x before we
did any canceling out. So hopefully that helps
clear things up a little bit.