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3.9 Derivative of Composite Functions

If a function u = φ(x) has, at some point x, a derivative u'ₓ = φ'(x), and the function y = F(u) has, at the corresponding value of u, the derivative yᵤ' = F'(u), then the composite function y = F[φ(x)] at the given point x also has a derivative, which is equal to yₓ' = Fᵤ'(u)φ'(x).

Proof.

For a definite value of x we will have, u = φ(x) and y = F(u)

Increase the value of the argument x by Δx,

u + Δu = φ(x + Δx),      y + Δy = F(u + Δu)

Thus for the increment Δx there corresponds a increment Δu to which corresponds an increment Δy.

As such Δu -> 0, Δy -> 0, and Δx -> 0

Hence, it is given

                 Δy
lim(Δx->0)---- = yᵤ'
                 Δu

By the definition of limit, from this relation we get,

Δy
---- = yᵤ' + α
Δu

Solve for Δy,

Δy = yᵤ'Δu + αΔu

Take the ratio of the increments Δy and Δx,

Δy         Δu       Δu
---- = yᵤ'---- + α----
Δx         Δx       Δx

As Δx -> 0, α ->0, so it is given that,

                Δu
lim(Δx->0)---- = uₓ', lim(Δx->0)α = 0
                Δx

Taking the limit of the ratio Δy/Δx,

                Δy
lim(Δx->0)---- = yᵤ'uₓ'
                Δy

Therefore, y'ₓ = yᵤ'uₓ'

Ex. y = sin(x²)

y = sinu, u = x²

yᵤ' = cosu, uₓ' = 2x

yₓ' = cos(u)2x = cos(x²)2x

3.10 Derivatives of y=tanx, y = cotx, y=ln|x|

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y = tanx : :
-----------------------------------------------------------

If y = tanx, then y' = 1/cos²x

Proof.

It is known that tanx = sinx/cosx

By the rule of differentiation of a fraction, we get

         (sinx)'cosx - sinx(cosx)'       cosxcosx + sinxsinx
y' = -------------------------------- = -------------------------
                        cos²x                              cos²x

                                                   cos²x + sin²x
                                               = ------------------
                                                          cos²x

By the trig identity that corresponds to Pythagorean theorem, sin²x + cos²x = 1

          1
y' = --------
       cos²x

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y = cotx : :
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If y = cotx, then y' = -1/sin²x

Proof.

It is known that cotx = cosx/sinx,

By the rule of differentiation of a fraction, we get,

       (cosx)'sinx - cosx(sinx)'    -sinxsinx - cosxcosx
y' = ---------------------------- = --------------------------
                     sin²x                              sin²x

                                              -(sin²x + cos²x)
                                           = -------------------
                                                       sin²x

By trig identity that corresponds to Pythagorean theorem sin²x + cos²x = 1, we get,

         -1
y' = -------
       sin²x

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y = ln|x| : :
-----------------------------------------------------------

If y = ln|x|, then y' = 1/x

Proof.

Let x > 0, then |x| = x, ln|x| = x,

therefore, y' = 1/x

Let x < 0, then |x| = -x and ln|x| = ln(-x)

y = ln(u) where u = -x

Then yₓ' = yᵤ'uₓ' = (1/u)(-1) = (-1)/(-x) = 1/x

hence, yₓ' = 1/x

3.11 An Implicit Function and its Derivative

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Implicit Functions : :
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Let the values of two variables x and y be related by some equation F(x,y)=0.

If the function y = f(x), defined on some interval (a,b), is such that F(x,y) = 0 becomes an identity in x when the expression f(x) is substituted for y, the function y = f(x) is called an Implicit Function.

While some functions, most that we are used to in fact, are defined explicitly in the form y = f(x), many are also defined implicitly via an equation F(x,y) = 0.

Note that implicit or explicit functions are not fundamentally different, they are merely functions defined in a specific context.

Consider x² + y² - a² = 0, we solve this equation for y to derive the implicit functions,

y = √[a² - x²]
y = -√[a² - x²]

We can check if these are in fact implicitly defined functions of the equation by plugging one back into the equation

x² + √[a² - x²]² - a² = 0

=> x² + a² - x² - a² = 0
=> (x² - x²) + (a² - a²) = 0

=> 0 = 0

Above we have converted an implicit form of a function into its explicit form, however not all implicit functions can
be translated into explicit functions.

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Derivative of Implicit Functions : :
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Again consider x² + y² - a² = 0

We differentiate all terms in the equation in respect to x, given a² is a constant it evaluates to zero,

2x(dx/dx) + 2y(dy/dx) - 0 = 2x(1) + 2y(dy/dx) = 0

Now we solve for dy/dx,

2x + 2y(dy/dx) = 0 => 2y(dy/dx) = -2x

=> dy/dx = -2x/2y
=> dy/dx = -x/y

Recall the explicit function is y = √[a² - x²] whose derivative is dy/dx = -x/√[a² - x²] = -x/y

Now consider y⁶ - y - x² = 0, this implicit function cannot be defined explicitly, we can still take its derivative,

6y⁵(dy/dx) - 1(dy/dx) - 2x(dx/dx) = 0 => 6y⁵(dy/dx) - dy/dx = 2x

=> dy/dx(6y⁵ - 1) = 2x
=> dy/dx = 2x/(6y⁵ - 1)

3.12 Derivatives of Power Functions, Exponential Functions, and Composite Exponential Functions

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y = xⁿ, where n ∈ R : :
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If y = xⁿ , then y' = nxⁿ⁻¹

Proof.

Let x > 0.

Taking the log on both sides of the function,

lny = lnxⁿ = nlnx

Differentiate in respect to x,

y'        1
--- = n---
y         x
               1
-> y' = yn--- = xⁿnx⁻¹ = nxⁿ⁻¹
               x

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y = aˣ where a > 0 : :
-----------------------------------------------------------

If y = aˣ, then y' = aˣlna

Proof.

Take the log of both sides of the function,

lny = lnaˣ = xlna

Differentiate the equality in respect to x,

 y'
--- = lna ---> y' = ylna = aˣlna
 y

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y = uᵛ : :
-----------------------------------------------------------

If y = uᵛ, then vuᵛ⁻¹u' + uᵛv'lnu

Proof.

Take the logarithm of the function,

lny = lnuᵛ = vlnu

Differentiate,

y'        1
--- = v---u' + v'lnu
y         u

u'      
y' = y(v(u/u') + v'lnu)
            u
          
y' = yv(u'/u) + yv'lnu = vuᵛ⁻¹u' + uᵛv'lnu

Adelaide said: 

3.12 Derivatives of Power Functions, Exponential Functions, and Composite Exponential Functions

-----------------------------------------------------------
y = xⁿ, where n ∈ R : :
-----------------------------------------------------------

If y = xⁿ , then y' = nxⁿ⁻¹

Proof.

Let x > 0.

Taking the log on both sides of the function,

lny = lnxⁿ = nlnx

Differentiate in respect to x,

y'        1
--- = n---
y         x
               1
-> y' = yn--- = xⁿnx⁻¹ = nxⁿ⁻¹
               x

-----------------------------------------------------------
y = aˣ where a > 0 : :
-----------------------------------------------------------

If y = aˣ, then y' = aˣlna

Proof.

Take the log of both sides of the function,

lny = lnaˣ = xlna

Differentiate the equality in respect to x,

 y'
--- = lna ---> y' = ylna = aˣlna
 y

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y = uᵛ : :
-----------------------------------------------------------

If y = uᵛ, then vuᵛ⁻¹u' + uᵛv'lnu

Proof.

Take the logarithm of the function,

lny = lnuᵛ = vlnu

Differentiate,

y'        1
--- = v---u' + v'lnu
y         u

u'      
y' = y(v(u/u') + v'lnu)
            u
          
y' = yv(u'/u) + yv'lnu = vuᵛ⁻¹u' + uᵛv'lnu

Are you Alice?

3.13 Inverse Functions and their Derivatives

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Increasing and Decreasing Functions : :
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Consider a increasing function y = f(x) defined on some interval (a,b) where a < b.

Let f(a) = c and f(b) = d.

Consider two different values x₁ and x₂ in the interval (a, b).

From the definition of an increasing function it follows that if x₁ < x₂ and y₁ = f(x₁), y₂ = f(x₂), then y₁ < y₂.

The converse is also true, that is if y₁ < y₂ and y₁ = f(x₁), y₂ = f(x₂), then x₁ < x₂.

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Inverse Functions : :
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In such a case, we get x as a function of y, x = φ(y), which is called the inverse function of y = f(x).

Given x = φ(y) is inverse of y = f(x), y = f(x) is inverse of x = φ(y).

If an increasing or decreasing function y = f(x) is defined over an interval [a,b] where f(a) = c and f(b) = d, then the inverse function is defined and is continuous on the interval [c,d].

Ex. y = x³

y = x³ is increasing on the interval -∞ < x < ∞, as such its inverse must too be defined on the interval -∞ < y < ∞

Solving for x yield, x = ∛y

If a function y = f(x) is not strictly increasing or decreasing over an interval it can have more than one inverse function.

Ex. y = x²

The function y = x² is defined over the interval -∞ < x < ∞, but over this interval it is neither strictly increasing or decreasing, therefore it does not have an inverse function.

If we consider y = x² over the interval -∞ < x < 0, then the function is strictly decreasing and will have the inverse function x = -√y

If we consider y = x² over the interval 0 < x < ∞ then the function is strictly increasing and will have the inverse function x = √y

If the functions y = f(x) and x = φ(y) are reciprocals, their graphs are represented by a single curve. Graphs will be symmetric about the bisector of the first quadrantal angle.

Ex. y = e^x and inverse x = lny has an inverse y = lnx

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The Derivative and Inverse Functions : :
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If for the function y = f(x) there exists an inverse function x = φ(y) which at the point under consideration y has a nonzero derivative φ'(y), then at the corresponding point x the function y = f(x) has a derivative f'(x) = 1/φ'(y)

Proof.

Differentiate in respect to x the inverse function x = φ(y),

1 = φ'(y)y'ₓ

Solve for y'ₓ,

y'ₓ = 1/φ'(y)