IB Class Notes

Chain Rule

Aim: By the end of this note, you will be able to take a derivative of a function of a function.

  1. We now know how to differentiate \( y = ax^n , ~ y= sin x, ~ y =cos x, ~ y = tan x, ~ y =ln x, ~ y= log_b{x} \) and \( y = e^x \) but with our present knowledge differentiating
    \( y = sin 3x \) or \( y = (ln x)^3 \) represents a challenge.

  2. Many functions can be written as functions of some other functions, i.e \( y = g(f(x)) \).
    For example, \( y= (3x+4)^5 \) can be made up of two simple functions of \(x\), namely \( f(x)= 3x+4 \) and \(g(x)= x^5 \) .

  3. Derivative from the first principles suggests that if we want to find out the derivative of \( y=g \left( f(x) \right) \) then we need to consider how a small change in \( x \delta x \) would change \(y\).

  4. Here is a schematic diagram of a composite function:
    \[ [ x ] \to u=f(x) \to [ u ] \to y=g(u) \to [ y ] \] \[ \left[\delta x \right] \to u = f (\delta x ) \to [\delta u ] \to y = g( \delta u ) \to \left[ \delta y \right] \]

  5. The schematic diagram shows that a very small change in \( x \) (say \( \delta x) \) would lead to a small change in the function \( f \) , (say \( \delta u) \). This small change in function f in turn causes a small change in function \( y \). Thus, it is reasonable to guess that average rate of change in \( y \) due to \( x \) [i.e. \( \frac{dy}{dx} \) ] is the product of average rate of change in \( y \) due to \( u \) and average rate of change in \( u \) due to \( x \). \[\large \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

  6. Note that as \( \delta x \to 0 \) so will \( \delta u \to 0 \) since a change in function \( u \) is due to a change in input \( x \). Similarly as \( \delta u \to 0 \) then \( \delta y \to 0 \) because a change in function \( y \) is due to a change in input \( u \). Therfore as the limit of \( \delta x \to 0 \) (from the first principles) we obtain \[\large \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

    • A basic proof similar to that in your text book. Proof
    • HL students may want to try out the proof of chain rule from the first principle.

  7. Example 1
    Find the derivative of the following functions with respect to x.
    1. y= sin 2x
    2. y= tan 5x

    Solution
    1. Let u=2x and y =sin u
      du/dx = 2 & dy/du = cos u
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (cos u)(2) But u= 2x
      dy/dx = 2 cos 2x

    2. Let u=5x and y = tan u
      du/dx= 5 & dy/du= 1/cos2u
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (1/cos2u)(5) But u= 5x
      dy/dx = 5/cos2(5x)

  8. Example 2
    Differentiate the following with respect to x:
    1. y= (3x+4)5
    2. y= (ln x)3
    3. y= (cos x)7
    4. y = sin2x
    5. y= \( \sqrt(2x^6-6x) \)
    Solution:
    1. Let u = 3x + 4 and y= u5
      du/dx = 3 & dy/du =5u4
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (5u4)(3) But u = 3x + 4.
      dy/dx = 15(3x + 4)4

    2. Let u = ln x and y = u3
      du/dx = 1/x & dy/du =3u2
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (3u2)(1/x) But u = ln x
      dy/dx = [3 (ln x)2] / x

    3. Let u =cos x and y = u7
      du/dx= -sin x & dy/du = 7u6
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (-sin x)(7u6) But u= cos x
      dy/dx = -7(sin x)(cos x)6

    4. sin2x = (sin x)2
      Let u = sin x and y = u2
      du/dx = cos x & dy/du = 2u
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (2u)(cos x) But u = sin x
      dy/dx= 2sin x cos x

    5. Let u = 2x6-6x and y = u1/2
      du/dx= 12x5-6 & dy/du = 1/2u-1/2
      \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
      dy/dx = (12x5-6)(1/2u-1/2) But u= 2x6-6x
      \( \frac{dy}{dx} = \frac{1}{2} \frac{12x^5 - 6}{\sqrt{2x^6-6x}} \\ \frac{dy}{dx} = \frac{6x^5 - 3}{\sqrt{2x^6-6x}} \)

    Rule for function raised to the power of \( n \)
    If you study the examples above carefully you will realize that there is a quick rule of thumb for differentiating \[ \begin{align} y & = \left[ f(x) \right]^ n \\ \frac{dy}{dx} & = {n[f(x)]^{n-1}}{f ' (x) } \end{align} \] Here you can regard \( f ' (x) = \frac{dy}{dx} \) as an adjustment to our previous power rule.

  9. Example 3
    Find the first derivative of y = [cos(x) + x]4 with respect to x.
    Solution
    dy/dx = ( 4[cos(x) + x]3 ) d[cos(x) + x]/dx
    dy/dx = ( 4[cos(x) + x]3 ) (-sin(x) + 1)

  10. Example 4
    Find the derivative for the following functions with respect to x:
    1. y= e3x
    2. y= e(x2-3x)
    3. y= etan x
    Solution
    1. Let u = 3x and y= eu
      du/dx= 3 & dy/du = eu
      dy/dx = (dy/du)(du/dx)
      dy/dx = (eu)(3) But u= 3x
      dy/dx = 3e3x

    2. Let u =x2-3x and y = eu
      du/dx= 2x -3 & dy/du=y = eu
      dy/dx = (dy/du)(du/dx)
      dy/dx = (eu)(2x-3) But u= x2-3x
      dy/dx = (2x-3)(e(x2-3x))

    3. Let u = tan x and y = eu
      du/dx= 1/cos2x & dy/du = eu
      dy/dx = (dy/du)(du/dx)
      dy/dx = (eu)( 1/cos2x) But u= tan x
      dy=dx = [ etan x ]/cos2x

    Rule for exponential function
    Here is another quick rule of thumb:
    \[ \begin{align} y & = e^{f(x)} \\ \frac{dy}{dx} & = f ' (x) e^{f(x)} \end{align} \]

  11. Example 5
    Differentiate the following functions with respect to x:
    1. ln 7x
    2. ln (sin x)
    3. ln (3 - x2)
    4. ln ( \(\large \frac{5}{\sqrt{x}} \) ) [attempt to complete this yourself]
    Solution
    1. Let u=7x and y = ln u
      du/dx= 7 & dy/du = 1/u
      dy/dx = (dy/du)(du/dx)
      dy/dx = (1/u)(7) But u= 7x
      dy/dx = 7/7x
      [ dy/dx = 1/x ]

    2. Let u=sin x and y = ln u
      du/dx= cos x & dy/du = 1/u
      dy/dx = (dy/du)(du/dx)
      dy/dx = (1/u)(cos x) But u= sin x
      dy/dx = cos x/sin x
      [ dy/dx = cot x ]

    3. Let u=3-x2 and y = ln u
      du/dx= -2x & dy/du = 1/u
      dy/dx = (dy/du)(du/dx)
      dy/dx = (1/u)(-2x) But u=3-x2
      dy/dx = -2x/(3-x2)
      [ dy/dx = 2x/(x2-3) ]
  12. Rule for Logarithmic function
    \[\begin{align} y & = ln [ f(x) ] \\ \frac{dy}{dx} & = \frac{f'(x)}{f(x)} \end{align} \].

  13. Example 6
    The decay of a radioactive element can be modelled using
    y = 20e-3t2 where y is the mass in gramme and t is time in hour. What is the rate of change at t=2?

    Solution
    Let u = -3t2 and y = 20eu
    du/dt = -6t & dy/du =20eu
    dy/dt = (dy/du)(du/dt)
    dy/dt = (20eu)(-6t) But u=-3t2
    dy/dt = -120te-3t2
    At t=2, dy/dt = -120(2)e-3(2)2
    At t=2, dy/dt = -240e-12
    At t=2, dy/dt = -0.00147 gramme ; the negative sign indicates "decay."
  14. Example 7
    The radius r of the base of a cyclinder is increasing at a rate of 4 cm/s. Find the rate at which the volume of this cyclinder (V = πr2h) is increasing when the radius is 5 cm in term of h and π.


    Solution
    The question gives us dr/dt = 4 cm/s and like us to find dV/dt at r=5.
    dV/dt = dV/dr x dr/dt ; (using chain rule)
    So first we need to obtain dV/dr.
    dV/dr = 2πrh
    dV/dt = 2πrh x (4)
    At r=5, dV/dt = 2π(5)h x (4)
    At r=5, dV/dt = 40πh cm3/s
  15. Complete the following table.
    \( y \)
    sin (kx)
    cos (kx)
    tan (kx)
    ekx
    ln (kx)
    \(\large \frac{dy}{dx} \)
     
     
     
     
     

Exercises.

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