# 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) $\large \frac{dy}{dx}$

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