Differential Calculus - Chain Rule for Trigonometric Functions
Summary
TLDRThis video covers the chain rule in differential calculus, specifically focusing on trigonometric functions. It explains how to differentiate composite trigonometric functions like sine, cosine, tangent, secant, cotangent, and cosecant using the chain rule. The instructor walks through several examples, highlighting common mistakes and emphasizing the importance of parentheses in simplifying the results. By the end, viewers learn how to differentiate functions within functions, apply the chain rule, and simplify expressions accurately. The video provides clear explanations for students to better understand and apply these concepts.
Takeaways
- 📘 The video covers the chain rule for trigonometric functions in differential calculus.
- 📌 The derivative of sin(U) is equivalent to cos(U) multiplied by U'.
- 🔍 The derivative of tan(U) is sec^2(U) multiplied by U'.
- ✏️ The derivative of cos(U) is -sin(U) multiplied by U'.
- 📐 For cot(U), the derivative is -csc^2(U) multiplied by U'.
- 📈 The chain rule is applied by differentiating the outer function, then multiplying by the derivative of the inner function (U').
- 📝 Example 1: For y = sin(2x), the derivative is 2 cos(2x).
- 🔑 Example 2: For y = cos(x-1), the derivative is -sin(x-1).
- 📊 Example 3: For y = tan(3x), the derivative is 3 sec^2(3x).
- ✅ The key takeaway is understanding how to apply the chain rule effectively for differentiating composite trigonometric functions.
Q & A
What is the chain rule in the context of trigonometric functions?
-The chain rule for trigonometric functions involves differentiating the outer function normally and then multiplying by the derivative of the inner function (U), which is referred to as U prime (U').
What is the derivative of sin(U) using the chain rule?
-The derivative of sin(U) using the chain rule is cos(U) multiplied by U'.
How do you apply the chain rule to differentiate tan(U)?
-To differentiate tan(U) using the chain rule, the result is sec²(U) multiplied by U'.
What is the formula for differentiating sec(U) using the chain rule?
-The derivative of sec(U) using the chain rule is sec(U)tan(U) multiplied by U'.
What happens when you differentiate cos(U) using the chain rule?
-When differentiating cos(U) using the chain rule, the result is -sin(U) multiplied by U'.
How do you find the derivative of cot(U) using the chain rule?
-The derivative of cot(U) using the chain rule is -csc²(U) multiplied by U'.
What steps should be followed when applying the chain rule to differentiate y = sin(2x)?
-First, identify the inner function U = 2x. Then, differentiate sin(U) to get cos(2x) and multiply it by the derivative of U, which is 2. The final answer is 2cos(2x).
How do you apply the chain rule to y = cos(x - 1)?
-First, identify U = x - 1. Then, differentiate cos(U) to get -sin(x - 1) and multiply it by the derivative of U, which is 1. The final answer is -sin(x - 1).
What is the result of differentiating y = tan(3x) using the chain rule?
-First, identify U = 3x. Differentiate tan(U) to get sec²(3x) and multiply it by the derivative of U, which is 3. The final answer is 3sec²(3x).
How do you differentiate y = sec(5x) using the chain rule?
-First, identify U = 5x. Differentiate sec(U) to get sec(5x)tan(5x) and multiply by the derivative of U, which is 5. The final answer is 5sec(5x)tan(5x).
Outlines
📘 Introduction to Chain Rule for Trigonometric Functions
The paragraph introduces the concept of the chain rule in differential calculus, focusing on trigonometric functions. It lists the derivatives of the six trigonometric functions (sin, tan, sec, cos, cot, and csc) with respect to an internal function 'U'. The method involves differentiating the outer function normally and then multiplying by the derivative of the inner function (U'). The example provided is y = sin(2x), demonstrating the chain rule process to obtain the final derivative.
📊 Example: Differentiating a Composite Cosine Function
This paragraph provides a second example where y = cos(x - 1). The chain rule is applied to differentiate the cosine function, with U being (x - 1). The derivative of y is calculated using the formula for the derivative of cosine, resulting in the final answer. Special emphasis is placed on correctly applying parentheses to avoid mistakes when writing the final expression.
🧮 Example: Derivative of Tangent Function with Inner Function
This example deals with y = tan(3x). The paragraph explains how to identify U as 3x and then apply the chain rule, using the formula for the derivative of tangent (sec²U). The derivative of y is simplified to the final form, demonstrating the application of the chain rule in trigonometric functions. Attention is given to proper simplification of terms involving sec².
🧑🏫 Applying Sum and Difference Rule in Chain Rule
This section introduces the sum and difference rule in the context of chain rule differentiation, using the example y = csc(5x) + 5. The csc(5x) part is differentiated separately, and then the sum rule is applied. The final derivative involves multiplying the derivative of csc by cotangent and simplifying to get the final result. The section also reiterates the importance of using parentheses correctly.
📐 Chain Rule in Complex Cosine Function Differentiation
The paragraph focuses on examples involving powers of trigonometric functions, specifically dealing with y = cos(3x²). The chain rule is applied by differentiating the outer cosine function and then the inner 3x² function. The final result is simplified, demonstrating the step-by-step application of the rule. The concept of treating cosine of constants differently is also highlighted in this section.
Mindmap
Keywords
💡Chain Rule
💡Trigonometric Functions
💡Sine Function (sin)
💡Cosine Function (cos)
💡Tangent Function (tan)
💡Secant Function (sec)
💡Composite Function
💡U-Prime (U')
💡Negative Sine (-sin)
💡Sum and Difference Rule
Highlights
Introduction to the chain rule for trigonometric functions, focusing on differentiation in calculus.
The derivative of sin(u) is presented as sin(u) * u', emphasizing the chain rule.
For tan(u), the derivative is derived as sec^2(u) * u'.
For sec(u), the derivative is given as sec(u) * tan(u) * u'.
The derivative of cos(u) is expressed as -sin(u) * u', showing the application of the chain rule.
Derivatives of cot(u) and csc(u) are discussed, highlighting their negative derivatives.
In the first example, the chain rule is applied to sin(2x), and the derivative is computed as 2 * cos(2x).
In the second example, the chain rule is used for cos(x - 1), resulting in -sin(x - 1).
For the third example, tan(3x) is differentiated as sec^2(3x) * 3, emphasizing the chain rule.
An example involving cot(5x) + 5 demonstrates the use of sum and difference rules with chain rule application.
Clarification on the importance of open and close parentheses in differentiating trigonometric functions.
The chain rule applied to complex trigonometric functions like cos(3x^2), showing detailed differentiation steps.
Using constant multiple rule for functions like cos(3) * x^2, showing how constants are treated in differentiation.
Differentiation of square roots in trigonometric functions, such as sqrt(cos(x)), by rewriting them in exponent form.
Conclusion: Emphasizing the importance of open and close parentheses, constant multiple rule, and chain rule when differentiating trigonometric functions.
Transcripts
hello class for this
video chain rule for trigonometric
functions this is a topic in
differential calculus the chain rule
versions of the derivatives of the six
trigonometric functions are shown below
first one we have the derivative of sin
U is equivalent to cine u u Prime second
we have the derivative of tan U is
equivalent to Second s u u Prime third
we have the derivative of second U U is
equivalent to Second U tangent u u Prime
fourth one we have the derivative of
cosine U is equivalent to netive sin u u
Prime next we have the derivative of
cotangent U is equivalent to netive cant
s u u Prime and last one we have the
derivative of cant U is equivalent to
negative cyant ENT u u
[Music]
Prime chain rule trigonometric function
all we need to do is to differentiate
them normally and then ilag
derivative U which is our U primea we
are differentiating a function within
our
function on how to apply chain rule on
our trigonometric functions first one we
have y is equivalent to S of 2x so as
you can observe class we have another
function within our function because you
X is mying two so this is another
function within our trigonometric
function so we have our formula Kina the
derivative of our sin U is equivalent to
cine u u Prime so U is equivalent to 2x
next thing we will derive our U that is
our U Prime is equivalent to two because
if we are deriving X raed to one that is
equivalent to one so u u Prime so proed
in differentiating our y that is
equivalent to Y Prime equals sign ising
cosine so that is cosine of is
substitute that is 2x multipied by our U
Prime which is two so copy paste two so
different shaping all we need to do here
is to simplify that is y Prime is
equivalent to take note multiply because
this is a trigonometric function
so
is front that is two cosine of 2x so
simplify PA and I think Hindi so this
will be our final answer so class
pite to avoid confusion that is 2 cosine
of open parentheses of 2x close
parentheses emphasize 2x is cosine so
this is our final answer so
P for our second example we have y is
equivalent to cosine of open parentheses
of x - 1 closed parenthesis so we will
be using this formula we have derivative
of cosine U is equivalent to negative of
sin U U Prime first thing that we need
to do is to determine U which
is cosine function which is our xus
one and then derive u u Prime that is
equivalent to the derivative of x is 1
the derivative of1 is zero so u u Prime
so let us proceed in deriving our y that
is y Prime is equivalent
toag cosine U is negative sin U so
negative open parenthesis of s then U is
xus one so open parentheses x
-1 close parenthesis close parenthesis
next thing is U Prime so multiply U
Prime which is one so we're done in
differentiating all we need to do is to
simplify this one that is equivalent to
negative imultiply
s of
xus so this will be our final
answer simplify common mistake class Isa
open and close parenthesis for example
negative s of x -1 so if this one
isare this is not equal
because is sin and then is deded by
one
avidus open and
clthes open and cl parentheses emphasize
xus one is the function of our sign so
first
example operation between
so open and close
parenthesis second example is
Operation function so again
class Final Answer third example is y is
equivalent to tangent of 3x first thing
that we need to do is to
determine which
isent which is c3x next class is to
derive our U which is our U Prime the
derivative of 3x is simply three so we
will be using this formula the
derivative of tangent U is equivalent to
Second s u u Prime so derive that is y
Prime is equivalent to tan U is second
squar of our U which is our U is
3X 3X
open and close parenthesis multiplied by
our U Prime which is three so we're done
in differentiating all we need to do is
to simplify so take note Class B IM
multiply because we are dealing with
trigonometric
function
so second squar of 3x and this will be
our final answer so take note class open
and close
parentheses
oper but again class to avoid
mes open and
cl trigonometric function
so Final Answer letter D or our fourth
example we have the function of X is
equivalent to open parenthesis of cant
5X close parentheses + 5 so clarate so
we will be using sum and difference rule
we just need to differentiate them
individually so for our first
part formula which is the derivative of
cyant U is equivalent to negative open
parentheses of cyant U cotangent U close
parentheses U Prime so first thing that
we need to do is to identify our U that
is equivalent
to cose function which is 5X next class
the derivative of our U is equivalent to
5 so proceed put first part which is the
derivative of our cose 5x we have F
Prime of X is equivalent to the
derivative of cant 5x is equivalent to
negative open parenthesis of cose of u u
is 5X lagay
cotangent of U so U is another
5x close parenthesis close parenthesis
then U Prime not in class is five
so part proceed second part which is the
derivative of our five so as we all know
the derivative with respect to X of any
constant is equivalent to zero so this
is plus Z so we're done differentiating
all we need to do is to simplify so fime
of X is equivalent to again class
multiply is front so
isant of 5x cotangent of
5x and this will be our final answer so
P Wala open and close parenthesis okay
Lang but again to avoid
mistakes that
is5 cose of open parentheses 5x
multiplied by cotangent of open and Clos
parentheses 5x so this will also be our
final answer so class to avoid confusion
open and close parenthesis here are our
examples first one is y is equivalent to
cosine of 3x^2 second we have y is
equivalent to open parentheses cosine of
3 Clos parentheses x^2 third we have y
is equivalent to cosine open parentheses
of 3x close parentheses squar fourth we
have y is equivalent to cosine 2 x and
last one we have y is equivalent to the
square root of cosine of x soive class
and let us
determine is the derivative of our
cosine U is equivalent to netive sin u u
Prime let's start with our letter A that
is equivalent to cosine of open
parenthesis
3x s so same open and
clthes
3X and cl parenthesis which is
3x² and then U Prime is equivalent to 3
* 2 is 6 x then 2 - 1 is 1 so 6 x to 1
so apply class formula y Prime is
equivalent to negative open parenthesis
cosine is mag
s then copy paste C that is 3x squared
close parenthesis then multiplied by our
U Prime so to U Prime that is
6X last thing that we need to do is
simplify so again Class
B
so
6X s of
3x^2 and this will be our final answer
so open and close parentheses 3x2 that
is s of open parenthesis 3 x² closed
parenthesis final answer so how about if
separate cosine 3 then open and close
parenthesis tapos multipli by X so analy
class cosine of 3 is a constant
variable so we will just treat cosine 3
as a
constant we will just differentiate this
one using our constant multiple rule so
recall the derivative with respect to X
of a constant u n is equivalent to
constant n u n minus one so y Prime is
equivalent to our constant which is a
cosine of
3 multiplied by our n which is
squared multiplied by our U which is our
X then 2 - 1 is one so rearrange class
this is equivalent to 2x cosine of 3
this is our final answer so
again
is so final answer
so front for our letter C we have cosine
of open parenthesis 3x then close
parenthesis squared so squ
CL is
3x
soite that is equivalent to cosine of 3^
s will be 9 then X is squared so that is
equivalent to x^ 2 so cosine of 9 x^2 is
same cosine of open parenthesis 9 x^2
closed parenthesis so we will be using
this formula the derivative of cosine of
U is equivalent to Nega of s u u Prime
first thing that we need to do is to
identify our U which is equivalent to
which is 9 X2 next one derive u u Prime
is equivalent to 9ti by 2 is 18 and then
x 2 - 1 is 1 so raised to 1 so
gam constant multiple rule so class we
have our U we have our U Prime so
proceed to in differentiating that is y
Prime is equivalent
to that is negative then open
parenthesis cosine is magig s u so magig
s of 9 X2 close parenthesis U Prime
class is C
18 x so as I have said
earlier is ilagay front so that is
equivalent to
-8x sin of 9 X2 this will be our final
answer so para sure
P open and close parentheses 9x2 so s of
open parentheses of 9 x^2 close
parenthesis final answer
so
okay
pro d that is
yal that is just equivalent to open
parenthesis of cosine X Clos parentheses
squ Sil is squared so as per
observation is chain rule constant
multiple
rule mle
rule U Prime because we are
differentiating a function within a
function so identify First what is our c
that
isan Conant front which is exponent that
is cosine of x
proceed U Prime which is the derivative
of our U so as we all know the
derivative with respect to X of our
cosine of x is equivalent to negative s
of X so the derivative of our cosine of
x is negative sin of X last one is
exponent which is n that is equivalent
to
two u u Prime and N let's proceed in
differentiating the is equivalent
to formula constant is n is two U is
cosine of x and then n is 2us 1 is 1
then U Prime is
netive sin of X take note open and close
parenthesis
so so that is equivalent
to negative so Lang front
negative -2 cosine of x sin of
X and this will be our final answer
P we have NE -2 sin of X and cosine of x
that is just the same so that is
equivalent to -2 sin of X multiplied by
cosine of x so same
Lang last example class is y is
equivalent to the square root of our
cosine of x so we cannot differentiate
this immediately we need to rewrite this
one that is equivalent to open
parenthesis cosine of x then rais to the
power of 12 because as we all know is
power of one square root is two so we
will be using end root of X ra to m is
equivalent X to m / n
soite and proed
inting Rule Conant multiple rule which
is this one so
identify front next
one U which is exponent which is cosine
of X
proceed U Prime which is the derivative
of cosine
X which is netive
sinx last thing is our exponent which is
n that is equivalent to 12 so class we
have our constant u u Prime and our n so
proceed differentiating y Prime is
equivalent to constant is w and
is
12 U class is cosine of x and then
exponent is 12 minus one is
one2 calculator para sure and last thing
is your U Prime which is negative sin of
X so in differentiating last thing that
we need to do is simplify
soay negative it transfer s front
12 of cosine of x raised to the power
of2 multiplied by
sinx so Final Answer exponent negative
so it
transfera so in transferring class
always remember that X ra to m is
equivalent to 1 /x to the power of M so
transfer class
negative sin
xang that is sin x divided by our cosine
of x then raised to the power of POS 12
and then
class that is two so
adjust so as per
observation okay Lang radical sa Baba so
this is our final answer so class if
choices you just need to
rationalize okay final answer so that
ends our topic for this video class do
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