• Product rule
  • Quotient rule
  • Chain rule
  • Partial differentation
  • Differentials
  • Total differentials
  • Total derivative
  • Derivative of implicit functions
  • Inverse function rule

Warm-up exercises

Round 1

Calculate $\dfrac{dy}{dx}$ for the following functions:

🐑 $y = x^2$

🐑 $y = 5x^3 + 12$

🐑 $y = 5x^{-2}$

🐑 $y = \frac{7}{x}$

🐑 $y = 18\sqrt{x}$

Round 2

🐑 $y = \ln(x)$

🐑 $y = \sin(x)$

🐑 Given the total-cost function $C = Q^3 - 5Q^2 + 12Q + 75$, write out a variable-cost ($VC$) function. Find the derivative of the VC function, and intepret the economic meaning of that derivative.

🐑 Provide a mathematical proof for the general result that, given a linear average curve, the corresponding marginal curve must have the same vertical intercept but will be twice as steep as the average curve.

Product rule

${\dfrac{d}{dx}}[f(x)\cdot g(x)]= f(x)\cdot {\dfrac {d \ [g(x)]}{dx}}+g(x)\cdot {\dfrac{d \ [f(x)]}{dx}}$

$(f\cdot g)'=f'\cdot g+f\cdot g'$

Watch Khan Academy's video on the product rule and the proof.

How do you remember the product rule?


Calculate $f'(x)$ for the following functions:

🐑 $f(x) = 5x^4(3x - 7)$ by multiplying out first and by using the product rule.

🐑 $f(x) = (x^8 + 8)(x^6 + 11)$ by multiplying out first and by using the product rule.

🐑 $f(x) = (4x^2-3)(2x^5)$

🐑 $f(x) = 7x^9(3x^2-12)$

🐑 $f(x) = (2x^4+5)(3x^5-8)$

🐑 $f(x) = (3-12x^3)(5+4x^6)$

Quotient rule

$f(x) = \dfrac{g(x)}{h(x)}$

$f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$

Watch Khan Academy's video on the quotient rule and the how the quotient rule can be derived from the product and chain rule.

How do you remember the quotient rule?


🐑 $f(x) = \dfrac{10x^8 - 6x^7}{2x}$ first by simplifying and then using the quotient rule.

🐑 $f(x) = \dfrac{3x^8-4x^7}{4x^3}$

🐑 $f(x) = \dfrac{4x^5}{1-3x}$ and $(x \neq \frac{1}{3})$

🐑 $f(x) = \dfrac{15x^2}{2x^2+7x-3}$

🐑 $f(x) = \dfrac{6x-7}{8x-5}$

🐑 $f(x) = \dfrac{5x^2-9x+8}{x^2+1}$

Chain rule

If $F(x) = f(g(x))$ then $F'(x)=f'(g(x))g'(x).$

Or equivalently, ${\dfrac {dz}{dx}}={\dfrac {dz}{dy}}\cdot {\dfrac {dy}{dx}}$

Watch Khan Academy's video on the chain rule and the proof.

How do you remember the chain rule?

Remember the chain rule with the Matryoshka_doll. Each nested function is a doll.


🐑 $f(x) = (3x^4 + 5)^6$

🐑 $f(x) = (7x + 9)^2$

🐑 $f(x) = (x^2 + 3x - 1)^5$

🐑 $f(x) = -3(x^2 - 8x + 7)^4$

🐑 $f(x) = (\ln(2x^2 + 1))^4$

🐑 $f(x) = \sin(7x + \ln(2x^3 + \dfrac{14}{(x+2)^3}))$

So far we have been analysing functions of one variable, $x$. Now we consider functions of more than one variable:

$z = f(x,y)$

How do we evaluate the impact of a change in $x$ on $z$ ?

Partial differentiation

Everything is the same. Just treat the other variables (in this case $y$) as constants.

\begin{align*} \dfrac{\partial z}{\partial x} \end{align*}


🐑 $z = 8x^2 + 14xy + 5y^2$

🐑 $z = 4x^3 + 2x^2y-7y^5$

🐑 $z = 6w^3 + 4wx + 3x^2 - 7xy - 8y^2$

🐑 $z = 2w^2 + 8wxy - x^2 + y^3$

🐑 $z = 3x^2(5x + 7y)$

🐑 $z = (9x-4y)(12x+2y)$


If you looked it up, a derivative is defined as $f'(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}$ and so we we have been treating $f'(x)$ or $\frac{dy}{dx}$ as a single object.

We can also write $dy = f'(x)dx$ where $dy$ and $dx$ are the differentials.

See page 179 of Chiang and Wainwright more details.


Find the differential $dy$ for the following functions:

🐑 $y = (4x +3)(3x-8)$

🐑 $y = (11x+9)^3$

Total differentials

Now use differentials on functions of several variables.

For the function $z = f(x,y)$

$dz = \dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy$

Simply find the partial derivatives and substitute into the above formula.

See page 184 of Chiang and Wainwright for more details.

And here is a nice video that gives great example of how to use total differentials.


Find the total differential $dz = z_x dx + z_y dy$ for the following functions:

🐑 $z = 5x^3-12xy-6y^5$

🐑 $z = 3x^2(8x-7y)$

Total derivative

Consider the case where $z = f(x,y)$ and we wish to calculate $\dfrac{dz}{dx}$. Before, we calculated a partial derivative by treating $y$ as a constant. What if $y$ is also a function of $x$ so that it no longer makes sense to change $x$ but keep $y$ constant? We then use the total derivative:

$\dfrac{dz}{dx} = z_x + z_y \dfrac{dy}{dx}$

Draw a dependency diagram.

See page 189 of Chiang and Wainwright for more details.

This video shows how we can use the total differential to derive the total derivative.


Find the total derivative $\dfrac{dz}{dx} = z_x + z_y \dfrac{dy}{dx}$ for the following functions:

🐑 $z = 6x^2 + 15xy + 3y^2$ where $y=7x^2$

🐑 $z = (13x - 18y)^2$ where $y = x + 6$

Derivative of implicit functions (equations)

We have been working with functions of the form $y = f(x)$. What if we are given $g(x,y) = k$ and it is too difficult to convert the second form into the first?

Step 1: Differentiate both sides of the equation with respect to $x$ while treating $y$ as a function of $x$.

Step 2: Solve for $\dfrac{dy}{dx}$

Details on the derivation of this formula are on page 197 of Chiang and Wainwright.

Implicit differentiation is a little tricky. Khan Academy have a great video making it super simple and more advanced examples. You will need implicit differentiation in macro.


Use implicit differentiation to find the derivative $\dfrac{dy}{dx}$ for the following equations:

🐑 $4x^2 - y^3 = 97$

🐑 $3y^5 - 6y^4 + 5x^6 = 243$

🐑 $x^4y^6 = 89$

🐑 $2x^3 + 5xy + 6y^2 = 87$

🐑 $(2x^3+7y)^2 = x^5$

Implicit Function Rule

Now suppose we don't have an equation, but just the function $f(x,y)$. How do we calculate $\dfrac{dy}{dx}$?

How could we derive a rule from the total differential? 🐑 |🐑

In general, if an implicit function is defined in the neighbourhood of $x$, then $\dfrac{dy}{dx}=-\dfrac{f_x}{f_y}$

We could also this formula for equations of the form $g(x,y) = k$.


Use the implicit function rule to find $\dfrac{dy}{dx}$.

🐑 $f(x,y) = 3x^2 +2xy + 4y^3$

🐑 $f(x,y)=12x^5-2y$

🐑 $f(x,y)=7x^2+2xy^2+9y^4$

🐑 $f(x,y)=6x^3-5y$

Inverse function rule

If the function is a one-to-one mapping then

$\dfrac{dx}{dy} = \dfrac{1}{\dfrac{dy}{dx}}$

Watch Khan Academy's video on the inverse function rule.


🐑 Are the following functions strictly monotonic (one-to-one mapping)?

$y = -x^6 + 5$ for $x>0$

$y=4x^5 + x^3 + 3x$

For each strictly monotonic function, find $\dfrac{dx}{dy}$ by the inverse-function rule.


Exercise 1

Differentiate the following:

🐑 $f(x) = \dfrac{3x(2x-1)}{5x-2}$

🐑 $f(x) = 3x(4x-5)^2$

🐑 $f(x) = \dfrac{(8x-5)^3}{7x+4}$

Exercise 2

Find all the partial derivatives of:

🐑 $z = (2x^2+6y)(5x-3y^3)$

🐑 $z = (w-x-y)(3w+2x-4y)$

🐑 $z = \dfrac{5x}{6x-7y}$

🐑 $z = \dfrac{x^2-y^2}{3x+2y}$

Exercise 3

Find the total differential $dz = z_x dx + z_y dy$ for the following functions:

🐑 $z = 7x^2y^3$

🐑 $z = (5x^2+7y)(2x-4y^3)$

Find the total derivative $\dfrac{dz}{dw}$ for the following functions:

🐑 $z = 7x^2 + 4y^2$ where $x=5w$ and $y=4w$

🐑 $z = 10x^2 - 6xy - 12y^2$ where $x = 2w$ and $y = 3w$

Exercise 4

Use the implicit function rule to find $\dfrac{dy}{dx}$ and $\dfrac{dy}{dz}$.

🐑 $f(x,y,z) = x^2y^3 + z^2 + xyz$

🐑 $f(x,y,z) = x^3z^2 + y^3 + 4xyz$