Product rule, quotient rule and rates

What you will be able to do

• differentiate a product and a quotient;
• find the derivatives of $\sec x$, $\csc x$ and $\cot x$;
• use the derivative as a rate of change, with units.

A product is not the product of the derivatives

Take $u = x$ and $v = x^2$. Then $uv = x^3$ has derivative $3x^2$. But $u'v' = 1 \times 2x = 2x$. So the derivative of a product is not the product of the derivatives.

The product rule

If $u$ and $v$ are differentiable, then $(uv)' = u'v + uv'$.

Here is the idea. Think of a rectangle with sides $u$ and $v$. When $x$ grows by $h$, the side $u$ grows by $\Delta u$ and the side $v$ grows by $\Delta v$. The area gains two thin strips and one tiny corner:

Divide by $h$ and let $h \to 0$. The two strips give $u'v$ and $uv'$. The corner is so small that, after the division, its share vanishes.

Example. For $f(x) = x^2\sin x$, take $u = x^2$ and $v = \sin x$. Then $f'(x) = 2x\sin x + x^2\cos x$.

The quotient rule

If $v \ne 0$, then $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.

It follows from the product rule. Write $\dfrac{u}{v} = u \cdot v^{-1}$, with $\dfrac{d}{dx}v^{-1} = -v^{-2}v'$. The order on top matters: derivative of the top times the bottom, minus the top times the derivative of the bottom.

Example. For $g(x) = \dfrac{x^2 + 1}{x - 1}$:

The other three trigonometric functions

The quotient rule on $\tan x = \dfrac{\sin x}{\cos x}$ gives $\dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$, as in the last lesson. In the same way:

• $\dfrac{d}{dx}\sec x = \sec x\tan x$;
• $\dfrac{d}{dx}\csc x = -\csc x\cot x$;
• $\dfrac{d}{dx}\cot x = -\csc^2 x$.

The three that begin with "co" have a minus sign. You can check this by noting that $\cos x$, $\csc x$ and $\cot x$ fall while $\sin x$, $\sec x$ and $\tan x$ rise just after $x = 0$.

The derivative as a rate

If $y$ depends on $x$, then $\dfrac{dy}{dx}$ is the rate at which $y$ changes when $x$ changes. Its unit is the unit of $y$ per unit of $x$.

• Speed. If $s(t)$ is the distance in metres at time $t$ seconds, then $v = \dfrac{ds}{dt}$ is the speed in m/s. For $s = t^3 - 6t^2 + 9t$ we get $v = 3t^2 - 12t + 9$. At $t = 2$ s the speed is $-3$ m/s, so the body moves backwards. It stops at $t = 1$ s and $t = 3$ s.
• A circle. The area is $A = \pi r^2$. The rate with respect to the radius is $\dfrac{dA}{dr} = 2\pi r$, which is the circumference. At $r = 5$ cm it is $10\pi \approx 31.4\ \text{cm}^2$ per cm. A thin ring of width $1$ cm around the circle has about that area.
• A sphere. The volume is $V = \dfrac{4}{3}\pi r^3$, and $\dfrac{dV}{dr} = 4\pi r^2$, the surface area.

This idea has a whole chapter in Class 12, which also handles two quantities that both change with time.

Practise with new numbers

Try it in Python

The program checks the product rule and the quotient rule against a difference quotient at $x = 1.3$. It also compares the rate of change of the area of a circle with $2\pi r$.

Common mistakes

• Multiplying the derivatives. $(uv)' \ne u'v'$. Use $u'v + uv'$.
• Swapping the top of the quotient rule. It is $u'v - uv'$, not $uv' - u'v$.
• Forgetting to square the bottom. The bottom of the quotient rule is $v^2$.
• Leaving out the unit of a rate. The derivative of an area with respect to a radius is in square centimetres per centimetre.

Remember these

• $(uv)' = u'v + uv'$.
• $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
• $(\sec x)' = \sec x\tan x$, $(\csc x)' = -\csc x\cot x$, $(\cot x)' = -\csc^2 x$.
• $\dfrac{dy}{dx}$ is a rate: units of $y$ per unit of $x$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Differentiate $y = x^3\cos x$.

2. Differentiate $y = \dfrac{x}{x^2 + 1}$.

3. Show that $\dfrac{d}{dx}\cot x = -\csc^2 x$ by writing $\cot x = \dfrac{\cos x}{\sin x}$.

4. A body moves so that $s = t^3 - 6t^2 + 9t$ metres at time $t$ seconds. Find its speed at $t = 4$ s.

5. How fast does the area of a circle grow with its radius when $r = 4$ cm?

6. A student writes $\dfrac{d}{dx}(x\sin x) = 1\cdot\cos x$. What is wrong?

References

• OpenStax, Calculus Volume 1, Chapter 3, sections 3.3 and 3.4 on the differentiation rules and the derivative as a rate of change.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, the sections on the derivative of a product and of a quotient of two functions.