Chain Rule
Aim: By the end of this note, you will be able to take a derivative of a function of a function.
- 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. - 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 \(g(x)= x^5 \) . - 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 \(y\).
- Here is a schematic diagram of a composite function:
\[ [ x ] \to u=f(x) \to [ u ] \to y=g(u) \to [ y ] \] \[ \left[\delta x \right] \to u = f (\delta x ) \to [\delta u ] \to y = g( \delta u ) \to \left[ \delta y \right] \] - 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 \( x \). \[\large \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]
- 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.
- Example 1
Find the derivative of the following functions with respect to x.- y= sin 2x
- y= tan 5x
- 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 - Let u=5x and y = tan u
du/dx= 5 & dy/du= 1/cos^{2}u
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
dy/dx = (1/cos^{2}u)(5) But u= 5x
dy/dx = 5/cos^{2}(5x)
- Example 2
Differentiate the following with respect to x:- y= (3x+4)^{5}
- y= (ln x)^{3}
- y= (cos x)^{7}
- y = sin^{2}x
- y= \( \sqrt(2x^6-6x) \)
- Let u = 3x + 4 and y= u^{5}
du/dx = 3 & dy/du =5u^{4}
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
dy/dx = (5u^{4})(3) But u = 3x + 4.
dy/dx = 15(3x + 4)^{4} - Let u = ln x and y = u^{3}
du/dx = 1/x & dy/du =3u^{2}
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
dy/dx = (3u^{2})(1/x) But u = ln x
dy/dx = [3 (ln x)^{2}] / x - Let u =cos x and y = u^{7}
du/dx= -sin x & dy/du = 7u^{6}
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
dy/dx = (-sin x)(7u^{6}) But u= cos x
dy/dx = -7(sin x)(cos x)^{6} - sin^{2}x = (sin x)^{2}
Let u = sin x and y = u^{2}
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 - Let u = 2x^{6}-6x and y = u^{1/2}
du/dx= 12x^{5}-6 & dy/du = 1/2u^{-1/2}
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
dy/dx = (12x^{5}-6)(1/2u^{-1/2}) But u= 2x^{6}-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. - 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) - Example 4
Find the derivative for the following functions with respect to x:- y= e^{3x}
- y= e^{(x2-3x})
- y= e^{tan x}
- Let u = 3x and y= e^{u}
du/dx= 3 & dy/du = e^{u}
dy/dx = (dy/du)(du/dx)
dy/dx = (e^{u})(3) But u= 3x
dy/dx = 3e^{3x} - Let u =x^{2}-3x and y = e^{u}
du/dx= 2x -3 & dy/du=y = e^{u}
dy/dx = (dy/du)(du/dx)
dy/dx = (e^{u})(2x-3) But u= x^{2}-3x
dy/dx = (2x-3)(e^{(x2-3x)}) - Let u = tan x and y = e^{u}
du/dx= 1/cos^{2}x & dy/du = e^{u}
dy/dx = (dy/du)(du/dx)
dy/dx = (e^{u})( 1/cos^{2}x) But u= tan x
dy=dx = [ e^{tan x} ]/cos^{2}x
Rule for exponential function
Here is another quick rule of thumb:
\[ \begin{align} y & = e^{f(x)} \\ \frac{dy}{dx} & = f ' (x) e^{f(x)} \end{align} \] - Example 5
Differentiate the following functions with respect to x:- ln 7x
- ln (sin x)
- ln (3 - x^{2})
- ln ( \(\large \frac{5}{\sqrt{x}} \) ) [attempt to complete this yourself]
- 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 ] - 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 ] - Let u=3-x^{2} 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-x^{2}
dy/dx = -2x/(3-x^{2})
[ dy/dx = 2x/(x^{2}-3) ]
- 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 = -3t^{2} and y = 20e^{u}
du/dt = -6t & dy/du =20e^{u}
dy/dt = (dy/du)(du/dt)
dy/dt = (20e^{u})(-6t) But u=-3t^{2}
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." - 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 = πr^{2}h) 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 cm^{3}/s - Complete the following table.
\( y \) sin (kx)cos (kx)tan (kx)e^{kx}ln (kx)\(\large \frac{dy}{dx} \)
Rule for Logarithmic function
\[\begin{align} y & = ln [ f(x) ] \\
\frac{dy}{dx} & = \frac{f'(x)}{f(x)}
\end{align} \].