Is there a more detailed lesson on the derivative of `|x|`?

Note that f(x) = |x| is not differentiable at x=0 The easiest way to do this is by rewriting |x| |x| = √(x²) (this is the principal square root only, do not simplify at this point) d/dx √(x²) = d/dx (x²)^(½) = ½ (x²)^(−½) ∙ d/dx (x²) = ½ (x²)^(−½) ∙ (2x) Rearranging: = ½ (2x)∙ (x²)^(−½) =(x)∙ (x²)^(−½) = (x)÷(x²)^(−½) = x ÷ |x| This is a piecewise function:``` f'(x) = −1, for x < 0 1, for x > 0 f'(x) is undefined at x=0 ```

if u=f(x^2-y^2,2xy) , find (partial^2u/partial x^2) let r=x^2-y^2 t=2xy how can i get this partial ??

By partial you mean the partial derivative of u with respect to x ?

I want to how to differentiate :y=√cosx+√cosx+√cosx+....∞ prove dy∕dx= sinx/1-2y. I wanna know how to differentiate infinite series.?HELP!!

First check your final expression it needs parenthesis to be readable. Both cos(x) and sqrt(x) have infinite series representation. You can take the derivative into a series term by term. But it looks like you can rewrite your series using some form of e^(ln(g(x))) or the like where g(x) is your original series. I suggest rewriting using lim (n-> inf) g(x) Then use facts like e^a * e^b = e^(a+b) and ln(a* b* c*) = ln a + ln b + ln (c) and for any power series that you can write down taking the derivative of the term by term is equivalent to integrating the derivative (look in the section)" operations on power series"

In the Practice: Differentiate composite functions I have a problem. I can't understand when 'v' is the outside part and when 'u' is the inside part of the function. In every exercise they change and I am getting confused.

For that purpose there is a rule of thumb. *ILATE* _I - Inverse trigonometric(sin^-1 , cos^-1 , etc. L - Logarithmic (log) A - Arithmatic (x^n , n is a real no.) T - Trigonometric (sin, cos , etc.) E - Exponential ( a^x)_ The function coming first (from left to right ) is taken to be 'u' and other one is taken to be 'v'. Hope that helps :)

Main content

Class 12 math (India)

Course: Class 12 math (India) > Unit 5

Lesson 9: Chain rule

Chain rule overview

Google Classroom

A quick overview sheet of the chain rule.

Introduction

If f, left parenthesis, x, right parenthesis and g, left parenthesis, x, right parenthesis are two functions, for instance f, left parenthesis, x, right parenthesis, equals, x, squared and g, left parenthesis, x, right parenthesis, equals, sine, left parenthesis, x, right parenthesis, we know how to take the derivative of their sum:


Rule:	start fraction, d, divided by, d, x, end fraction, left parenthesis, f, left parenthesis, x, right parenthesis, plus, g, left parenthesis, x, right parenthesis, right parenthesis, equals, f, prime, left parenthesis, x, right parenthesis, plus, g, prime, left parenthesis, x, right parenthesis
Example:	start fraction, d, divided by, d, x, end fraction, left parenthesis, x, squared, plus, sine, left parenthesis, x, right parenthesis, right parenthesis, equals, 2, x, plus, cosine, left parenthesis, x, right parenthesis

We also know how to take the derivative of their product:


Rule:	start fraction, d, divided by, d, x, end fraction, left parenthesis, f, left parenthesis, x, right parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, equals, f, left parenthesis, x, right parenthesis, g, prime, left parenthesis, x, right parenthesis, plus, g, left parenthesis, x, right parenthesis, f, prime, left parenthesis, x, right parenthesis
Example:	start fraction, d, divided by, d, x, end fraction, left parenthesis, x, squared, dot, sine, left parenthesis, x, right parenthesis, right parenthesis, equals, x, squared, cosine, left parenthesis, x, right parenthesis, plus, sine, left parenthesis, x, right parenthesis, dot, 2, x

The chain rule now tells us how to take the derivative of their composition, meaning f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis:


Rule:	start color #0c7f99, start fraction, d, divided by, d, x, end fraction, left parenthesis, f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, equals, f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis, end color #0c7f99
Example:	start color #0c7f99, start fraction, d, divided by, d, x, end fraction, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, squared, equals, 2, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, cosine, left parenthesis, x, right parenthesis, end color #0c7f99

Intuition using fake algebra

Warning: The following section may induce headache or queasiness for readers who are sensitive to violent abuse of notation.

We tend to write functions and derivatives in terms of the variable x.

start fraction, d, divided by, d, x, end fraction, left parenthesis, x, squared, right parenthesis, equals, 2, x

But of course, we could use any other letter.

start fraction, d, divided by, d, start color #11accd, a, end color #11accd, end fraction, left parenthesis, start color #11accd, a, end color #11accd, squared, right parenthesis, equals, 2, start color #11accd, a, end color #11accd

What if we did something crazy, and replaced x with a function instead of another letter.

start fraction, d, divided by, d, left parenthesis, start color #11accd, sine, left parenthesis, x, right parenthesis, end color #11accd, right parenthesis, end fraction, left parenthesis, start color #11accd, sine, left parenthesis, x, right parenthesis, end color #11accd, right parenthesis, squared, equals, 2, start color #11accd, sine, left parenthesis, x, right parenthesis, end color #11accd

It's unclear exactly what this start fraction, d, divided by, d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, end fraction symbol would mean, but let's just go with it for a second. We can imagine multiplying it by start fraction, d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, divided by, d, x, end fraction to "cancel out" the d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis term:

start fraction, d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, squared, divided by, start cancel, d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, end cancel, end fraction, dot, start fraction, start cancel, d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, end cancel, divided by, d, x, end fraction, equals, start fraction, d, divided by, d, x, end fraction, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, squared

This is not really a mathematically legitimate thing to do, since these "d, x" and "d, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis" terms are not numbers or functions that we can cancel out. There are ways to make this more legitimate that involve some more advanced math, but for now you can think of it as a useful memory trick. The usefulness is that when we expand start fraction, d, divided by, d, x, end fraction, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, squared like this, we know what each individual term is, even if we don't know how to take the derivative of left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis, squared:

\begin{array}{r} \frac{d}{d x} (\sin (x))^{2} = \underset{\begin{array}{c} Imagine replacing x with \sin (x) in \frac{d (x^{2})}{d x} \end{array}}{\underset{⏟}{\frac{d (\sin (x))^{2}}{d (\sin (x))}}} \cdot \underset{\begin{array}{c} Ordinary derivative of \sin (x) \end{array}}{\underset{⏟}{\frac{d (\sin (x))}{d x}}} = 2 \sin (x) \cdot \cos (x) \end{array}

This trick looks particularly clean when we write it in the abstract, rather than with the specific case of x, squared and sine, left parenthesis, x, right parenthesis:

start box, start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, close bracket, equals, start fraction, d, f, divided by, d, g, end fraction, dot, start fraction, d, g, divided by, d, x, end fraction, end box

Example 1:

\begin{aligned} f(x) &= \sin(x^2) \quad \quad \small{\gray{\text{Function to differentiate}}} \\ & \\ u(x) &= x^2 \quad \quad \small{\gray{\text{Define $u(x)$ as inside function}}} \\ & \\ f(x) &= \sin(u) \quad \quad \small{\gray{\text{Express $f(x)$ in terms of $u(x)$}}} \\ & \\ \frac{df}{dx} &= \frac{df}{du} \cdot \frac {du}{dx} \quad \quad \small{\gray{\text{Express chain rule applicable here}}} \\ & \\ \frac{df}{dx} &= \frac{d}{du}\sin(u) \cdot\frac {d}{dx}(x^2) \quad \quad \small{\gray{\text{Substitute in $f(u)$ and $u(x)$}}} \\ & \\ \frac{df}{dx} &= \cos(u) \cdot 2x \quad \quad \small{\gray{\text{Evaluate derivatives}}} \\ & \\ \frac{df}{dx} &= \cos(x^2)(2x) \quad \quad \small{\gray{\text{Substitute $u$ in terms of $x$.}}} \\ & \\ \end{aligned}

Example 2:

One pretty cool thing we can now do is find the derivative of the absolute value function vertical bar, x, vertical bar, which can be defined as square root of, x, squared, end square root. For example, vertical bar, minus, 5, vertical bar, equals, square root of, left parenthesis, minus, 5, right parenthesis, squared, end square root, equals, square root of, 25, end square root, equals, 5

\begin{aligned} f(x) &= \left|x\right| \quad \quad \small{\gray{\text{Function to differentiate}}} \\ & \\ f(x) &= \sqrt{x^2} \quad \quad \small{\gray{\text{Equivalent function}}} \\ & \\ u(x) &= x^2 \quad \quad \small{\gray{\text{Define $u(x)$ as inside function}}} \\ & \\ f(x) &= [u(x)]^\frac{1}{2} \quad \quad \small{\gray{\text{Express $f(x)$ in terms of $u(x)$}}} \\ & \\ \frac{df}{dx} &= \frac{df}{du} \cdot \frac {du}{dx} \quad \quad \small{\gray{\text{Express chain rule applicable here}}} \\ & \\ \frac{df}{dx} &= \frac{d}{du}u^\frac{1}{2} \cdot\frac {d}{dx}(x^2) \quad \quad \small{\gray{\text{Substitute in $f(u)$ and $u(x)$}}} \\ & \\ \frac{df}{dx} &= \frac{1}{2}u^{-\frac{1}{2}} \cdot 2x \quad \quad \small{\gray{\text{Compute derivatives with power rule}}} \\ & \\ \frac{df}{dx} &= \frac{1}{2}\left(x^2\right)^{-\frac{1}{2}} \cdot 2x \quad \quad \small{\gray{\text{Substitute $u(x)$ back in terms of $x$}}} \\ & \\ \frac{df}{dx} &= \frac{x}{\sqrt{x^2}} \quad \quad \small{\gray{\text{Simplify}}} \\ & \\ \frac{df}{dx} &= \frac{x}{\left|x\right|} \quad \quad \small{\gray{\text{Express $\sqrt{x^2}$ as absolute value.}}} \\ & \\ \end{aligned}

Arbitrarily long composition

The chain rule can apply to composing multiple functions, not just two. For example, suppose A, left parenthesis, x, right parenthesis, B, left parenthesis, x, right parenthesis, C, left parenthesis, x, right parenthesis and D, left parenthesis, x, right parenthesis are four different functions, and define f to be their composition:

f, left parenthesis, x, right parenthesis, equals, A, left parenthesis, B, left parenthesis, C, left parenthesis, D, left parenthesis, x, right parenthesis, right parenthesis, right parenthesis, right parenthesis

Using the start fraction, d, f, divided by, d, x, end fraction notation for the derivative, we can apply the chain rule as:

start fraction, d, f, divided by, d, x, end fraction, equals, start fraction, d, divided by, d, x, end fraction, A, left parenthesis, B, left parenthesis, C, left parenthesis, D, left parenthesis, x, right parenthesis, right parenthesis, right parenthesis, equals, start fraction, d, A, divided by, d, B, end fraction, dot, start fraction, d, B, divided by, d, C, end fraction, dot, start fraction, d, C, divided by, d, D, end fraction, dot, start fraction, d, D, divided by, d, x, end fraction

Using f, prime notation, we get more of a snowman aesthetic:

f, prime, left parenthesis, x, right parenthesis, equals, A, prime, left parenthesis, B, left parenthesis, C, left parenthesis, D, left parenthesis, x, right parenthesis, right parenthesis, right parenthesis, right parenthesis, dot, B, prime, left parenthesis, C, left parenthesis, D, left parenthesis, x, right parenthesis, right parenthesis, right parenthesis, dot, C, prime, left parenthesis, D, left parenthesis, x, right parenthesis, right parenthesis, dot, D, prime, left parenthesis, x, right parenthesis

Example 4:

Suppose f, left parenthesis, x, right parenthesis, equals, sine, left parenthesis, e, start superscript, x, squared, plus, x, end superscript, right parenthesis.

We think of f as being the composition of

\begin{aligned} A(x) &= \blueE{\sin(x)}\\ B(x) &= \greenE{e^x} \\ C(x) &= \redE{x^2 + x} \end{aligned}

Where the derivative of each function is

\begin{aligned} A'(x) &= \blueE{\cos(x)}\\ B'(x) &= \greenE{e^x} \\ C'(x) &= \redE{2x + 1} \end{aligned}

According to the chain rule, the derivative of the composition is

\begin{aligned} f'(x) &= A'(B(C(x))) \cdot B'(C(x)) \cdot C'(x) \\ & \\ &= \boxed{\large \blueD{\cos}(e^{x^2 + x}) \cdot \greenD{e}^{x^2 + x} \cdot \redD{(2x + 1)}} \end{aligned}

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Mevlan Isufi
Posted 8 years ago. Direct link to Mevlan Isufi's post “I just finished this tuto...”
I just finished this tutorial about the Chain Rule by watching all videos and solving all given problems; and now im confused at this overview. At the very first Introduction they mention "We also know how to take the derivative of their product:". I watched all videos and solved tens of problems and never heard or faced any problem like that. Looks like they will introduce it in next tutorials about "Taking Derivatives" or what? Please HELP !
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(8 votes)
Answer
- Joe Gaspar
  Posted 7 years ago. Direct link to Joe Gaspar's post “Following the videos in o...”
  Following the videos in order as presented on this website does in fact make that statement confusing.
  
  The rule in question is the "Power Rule" and those videos are presented in a section or two after the chain rule videos.
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  (3 votes)
VikingTheDude
Posted 8 years ago. Direct link to VikingTheDude's post “Is there a more detailed ...”
Is there a more detailed lesson on the derivative of |x|?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Just Keith
  Posted 8 years ago. Direct link to Just Keith's post “Note that f(x) = |x| is n...”
  Note that f(x) = |x| is not differentiable at x=0
  The easiest way to do this is by rewriting |x|
  |x| = √(x²) (this is the principal square root only, do not simplify at this point)
  d/dx √(x²)
  = d/dx (x²)^(½)
  = ½ (x²)^(−½) ∙ d/dx (x²)
  = ½ (x²)^(−½) ∙ (2x)
  Rearranging:
  = ½ (2x)∙ (x²)^(−½)
  =(x)∙ (x²)^(−½)
  = (x)÷(x²)^(−½)
  = x ÷ |x|
  This is a piecewise function:
  f'(x) = −1, for x < 0 1, for x > 0 f'(x) is undefined at x=0
  Comment on Just Keith's post “Note that f(x) = |x| is n...”
  (15 votes)
klimax
Posted 8 years ago. Direct link to klimax's post “On the last example, we c...”
On the last example, we can treat x^2+x as a function. But why we cannot treat 2 in x^2 as a function?
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(6 votes)
Answer
- Grant
  Posted 8 years ago. Direct link to Grant's post “While you can treat "2" a...”
  While you can treat "2" as a function, namely the constant function f(x) = 2 that outputs 2 for all inputs, raising one function to the power of another like f(x)^g(x) is different from composing functions, as in f(g(x)). For example, the chain rule cannot be used to take the derivative of x^x.
  Button navigates to signup page
  (1 vote)
Mary Shultz
Posted 8 years ago. Direct link to Mary Shultz's post “It appears that examples ...”
It appears that examples one and two are the same.
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(5 votes)
Answer
- Tanner Thygerson
  Posted 7 years ago. Direct link to Tanner Thygerson's post “The two problems are diff...”
  The two problems are different because of the position of the ^2 symbol. The sin function is being squared in example 1, and x is being squared in example 2.
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  (1 vote)
miles marcuson
Posted 6 years ago. Direct link to miles marcuson's post “I'm appalled by this viol...”
I'm appalled by this violent abuse of notation. What did that derivative ever do to you?
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(5 votes)
Answer
MaHmoud ImBabi
Posted 8 years ago. Direct link to MaHmoud ImBabi's post “if u=f(x^2-y^2,2xy) , fin...”
if u=f(x^2-y^2,2xy) , find (partial^2u/partial x^2)
let r=x^2-y^2
t=2xy
how can i get this partial ??
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(2 votes)
Answer
- Ahmad Youssef
  Posted 8 years ago. Direct link to Ahmad Youssef's post “By partial you mean the p...”
  By partial you mean the partial derivative of u with respect to x ?
  Button navigates to signup page
  (2 votes)
DeepShankarPratap
Posted 8 years ago. Direct link to DeepShankarPratap's post “I want to how to differen...”
I want to how to differentiate :y=√cosx+√cosx+√cosx+....∞ prove dy∕dx= sinx/1-2y.
I wanna know how to differentiate infinite series.?HELP!!
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Don Eckford
  Posted 8 years ago. Direct link to Don Eckford's post “First check your final ex...”
  First check your final expression it needs parenthesis to be readable.
  Both cos(x) and sqrt(x) have infinite series representation.
  You can take the derivative into a series term by term.
  But it looks like you can rewrite your series using some form of e^(ln(g(x))) or the like where g(x) is your original series.
  I suggest rewriting using lim (n-> inf) g(x)
  Then use facts like e^a * e^b = e^(a+b)
  and ln(a* b* c*) = ln a + ln b + ln (c)
  and for any power series that you can write down taking the derivative of the term by term
  is equivalent to integrating the derivative (look in the section)" operations on power series"
  Button navigates to signup page
  (4 votes)
Борислав Николов
Posted 7 years ago. Direct link to Борислав Николов's post “In the Practice: Differen...”
In the Practice: Differentiate composite functions I have a problem. I can't understand when 'v' is the outside part and when 'u' is the inside part of the function. In every exercise they change and I am getting confused.
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(1 vote)
Answer
- Vishal Pal
  Posted 7 years ago. Direct link to Vishal Pal's post “For that purpose there is...”
  For that purpose there is a rule of thumb.
  ILATE
  
  I - Inverse trigonometric(sin^-1 , cos^-1 , etc.
  L - Logarithmic (log)
  A - Arithmatic (x^n , n is a real no.)
  T - Trigonometric (sin, cos , etc.)
  E - Exponential ( a^x)
  
  The function coming first (from left to right ) is taken to be 'u' and other one is taken to be 'v'.
  
  Hope that helps :)
  Button navigates to signup page
  (3 votes)
Martin Xia
Posted 8 years ago. Direct link to Martin Xia's post “A(e^(x^2+x))=sin(e^(x^2+x...”
A(e^(x^2+x))=sin(e^(x^2+x))
B(e^x)=e^(x^2+x)=(e^x)^(x+1)
C(x)=e^x
then,
f'(x)=cos(e^(x^2+x))*e^(x^2+x)*(x+1)
Why?
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(2 votes)
Answer
wengi
Posted 7 years ago. Direct link to wengi's post “Just a quick question abo...”
Just a quick question about something barely related to this subject.
In example 3, at some point we get
f(x) = √(x²) = (x²)^½

Wouldn't we then be able to expand (x²)^½ in that way :

(x²)^½ --> x^(2/2) --> x^1 --> x

Somewhere along the way the |x| is turned into a simple x.

What is wrong with my logic ?
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(2 votes)
Answer
- Vu
  Posted 7 years ago. Direct link to Vu's post “Normally we could and wou...”
  Normally we could and would do that. But since f(x)=|x| is a piecewise function, f(x)=|x| {-x for x<0 ; x for x>0}, we want to keep it √(x²) as a one-piece function for our purpose of taking its derivative.
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  (1 vote)