Calculus · Paper 1
Differentiation — Part 2
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Section 1 of 11
Standard Derivatives You'll Need
Part 1 was all about powers of $x$. From here on, you'll bump into $\sin x$, $\cos x$, $\ln x$ and $e^{x}$ — usually mixed in with powers. Here are the standard derivatives you need.
Must learn — straight from the tables
$y = \sin x \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \cos x$
$y = \cos x \;\;\Rightarrow\;\; \dfrac{dy}{dx} = -\sin x$
$y = \ln x \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{1}{x}$
$y = e^{x} \;\;\Rightarrow\;\; \dfrac{dy}{dx} = e^{x}$
Note: $\ln x$ is just $\log_{e} x$ — log base $e$. It's on the maths tables under "natural log". And $e^{x}$ is the only function whose derivative is itself — neat.
Section 2 of 11
Which Rule? The $x = 10$ Trick
Three new rules are coming: Product, Quotient, Chain. Before you can use them, you need to know which one a question is asking for.
My trick: sub in $x = 10$ and watch what the very last operation is.
(i) $y = \dfrac{2x+1}{3x+5}$
At $x = 10$:
Top: $\;2(10) + 1 = 21$
Bottom: $\;3(10) + 5 = 35$
$y = \dfrac{21}{35}$ — two answers divided
Divide $\Rightarrow$ Quotient Rule
(ii) $y = x^{2} \sin x$
At $x = 10$:
$x^{2} = 100$
$\sin 10 \approx 0.173$ (radians)
$y = (100)(0.173)$ — two answers multiplied
Multiply $\Rightarrow$ Product Rule
(iii) $y = (2x+1)^{3}$
At $x = 10$:
$2(10) + 1 = 21$
$y = 21^{3} = 9261$ — the cube of $21$
"Of" $\Rightarrow$ Chain Rule
(iv) $y = \sin(3x+1)$
At $x = 10$:
$3(10) + 1 = 31$
$y = \sin 31$ — the sine of $31$
"Of" $\Rightarrow$ Chain Rule
Must learn — the identifier
By (multiplied) $\Rightarrow$ Product
Divide by $\Rightarrow$ Quotient
Of $\Rightarrow$ Chain
You try
Which rule for $\;y = (5x+3)^{4}$? (Sub in $x = 10$ to check.)
$5(10)+3 = 53$, then raise it to the 4. One operation "of" the bracket.
At $x = 10$: $\;5(10)+3 = 53$
$y = 53^{4}$ — the 4th power of $53$
Chain Rule
Chain Rule
You try
Which rule for $\;y = x^{3} \ln x$?
Two pieces: $x^{3}$ and $\ln x$. What operation joins them?
At $x = 10$: $\;x^{3} = 1000$, $\;\ln 10 \approx 2.30$
$y = (1000)(2.30)$ — two answers multiplied
Product Rule
Product Rule
You try
Which rule for $\;y = \dfrac{7x-3}{5-x}$?
Top and bottom — what joins them?
Top: $7(10)-3 = 67$. Bottom: $5-10 = -5$.
$y = \dfrac{67}{-5}$ — divided
Quotient Rule
Quotient Rule
Section 3 of 11
The Product Rule
When $y$ is the product of two functions — two pieces multiplied — call the first piece $u$ and the second piece $v$.
Product Rule — Must learn
$y = u\,v \;\;\Rightarrow\;\; \dfrac{dy}{dx} = u\,\dfrac{dv}{dx} \,+\, v\,\dfrac{du}{dx}$
Mnemonic: Leave Differ + Diff Leave
Leave the first, differentiate the second + differentiate the first, leave the second.
(i) $y = x^{2} \sin x$
Two pieces multiplied — Product Rule.
$u = x^{2}$ $v = \sin x$
$\dfrac{du}{dx} = 2x$ $\dfrac{dv}{dx} = \cos x$
$\dfrac{dy}{dx} = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx}$
$\dfrac{dy}{dx} = x^{2} \cos x + 2x \sin x$
(ii) $y = x \ln x$
$u = x$ $v = \ln x$
$\dfrac{du}{dx} = 1$ $\dfrac{dv}{dx} = \dfrac{1}{x}$
$\dfrac{dy}{dx} = x \cdot \dfrac{1}{x} + \ln x \cdot 1$
$\dfrac{dy}{dx} = 1 + \ln x$
(iii) $y = x^{5} \cos x$
$u = x^{5}$ $v = \cos x$
$\dfrac{du}{dx} = 5x^{4}$ $\dfrac{dv}{dx} = -\sin x$
$\dfrac{dy}{dx} = x^{5}(-\sin x) + \cos x \cdot 5x^{4}$
$\dfrac{dy}{dx} = -x^{5} \sin x + 5x^{4} \cos x$
$\dfrac{dy}{dx} = x^{4}\bigl(5\cos x - x \sin x\bigr)$ (factor out $x^{4}$)
Watch the minus sign on $\cos x$ — the derivative of $\cos$ is $-\sin$. Lose that minus and you'll lose marks.
(iv) $y = (3x+1) e^{x}$
$u = 3x+1$ $v = e^{x}$
$\dfrac{du}{dx} = 3$ $\dfrac{dv}{dx} = e^{x}$
$\dfrac{dy}{dx} = (3x+1) e^{x} + e^{x} \cdot 3$
$\dfrac{dy}{dx} = (3x+1) e^{x} + 3 e^{x}$
(v) $y = e^{x} \ln x$
$u = e^{x}$ $v = \ln x$
$\dfrac{du}{dx} = e^{x}$ $\dfrac{dv}{dx} = \dfrac{1}{x}$
$\dfrac{dy}{dx} = e^{x} \cdot \dfrac{1}{x} + \ln x \cdot e^{x}$
$\dfrac{dy}{dx} = \dfrac{e^{x}}{x} + e^{x} \ln x$
Section 4 of 11
Product Rule — Your Turn
You try
Differentiate $\;y = x^{3} \sin x$.
Two pieces multiplied. $u = x^{3}$, $v = \sin x$. Leave Differ $+$ Diff Leave.
$u = x^{3}$, $\;\;v = \sin x$
$\dfrac{du}{dx} = 3x^{2}$, $\;\;\dfrac{dv}{dx} = \cos x$
$\dfrac{dy}{dx} = x^{3} \cos x + \sin x \cdot 3x^{2}$
$\dfrac{dy}{dx} = x^{3} \cos x + 3x^{2} \sin x$
$x^{3} \cos x + 3x^{2} \sin x$
You try
Differentiate $\;y = x^{2} \ln x$.
$u = x^{2}$, $v = \ln x$. The $x^{2} \cdot \frac{1}{x}$ piece will simplify to $x$.
$u = x^{2}$, $\;\;v = \ln x$
$\dfrac{du}{dx} = 2x$, $\;\;\dfrac{dv}{dx} = \dfrac{1}{x}$
$\dfrac{dy}{dx} = x^{2} \cdot \dfrac{1}{x} + \ln x \cdot 2x$
$\dfrac{dy}{dx} = x + 2x \ln x$
$x + 2x \ln x$
You try
Differentiate $\;y = x \cos x$.
$u = x$, $v = \cos x$. Remember $\dfrac{d}{dx}(\cos x) = -\sin x$.
$u = x$, $\;\;v = \cos x$
$\dfrac{du}{dx} = 1$, $\;\;\dfrac{dv}{dx} = -\sin x$
$\dfrac{dy}{dx} = x(-\sin x) + \cos x \cdot 1$
$\dfrac{dy}{dx} = -x \sin x + \cos x$
$\cos x - x \sin x$
You try
Differentiate $\;y = (2x+5) e^{x}$.
$u = 2x+5$, $v = e^{x}$. The $e^{x}$ is the easy one — differentiating it leaves it the same.
$u = 2x+5$, $\;\;v = e^{x}$
$\dfrac{du}{dx} = 2$, $\;\;\dfrac{dv}{dx} = e^{x}$
$\dfrac{dy}{dx} = (2x+5) e^{x} + e^{x} \cdot 2$
$\dfrac{dy}{dx} = (2x+5) e^{x} + 2 e^{x}$
$(2x+5)e^{x} + 2e^{x}$
Section 5 of 11
The Quotient Rule
When $y$ is one function divided by another, top is $u$, bottom is $v$.
Quotient Rule — Must learn
$y = \dfrac{u}{v} \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{v\,\dfrac{du}{dx} \,-\, u\,\dfrac{dv}{dx}}{v^{2}}$
Order matters! $v \cdot du$ first, then minus $u \cdot dv$. Then divide by $v^{2}$ (the bottom squared).
(i) $y = \dfrac{3x+1}{7x+5}$ (where $x$ ≠ $-\tfrac{5}{7}$)
$u = 3x+1$ $v = 7x+5$
$\dfrac{du}{dx} = 3$ $\dfrac{dv}{dx} = 7$
$\dfrac{dy}{dx} = \dfrac{v\,\dfrac{du}{dx} - u\,\dfrac{dv}{dx}}{v^{2}}$
$\dfrac{dy}{dx} = \dfrac{3(7x+5) - 7(3x+1)}{(7x+5)^{2}}$
$\dfrac{dy}{dx} = \dfrac{21x + 15 - 21x - 7}{(7x+5)^{2}}$
$\dfrac{dy}{dx} = \dfrac{8}{(7x+5)^{2}}$
Notice $\dfrac{8}{(7x+5)^{2}} > 0$ for every $x$ in the domain — so this function is always increasing. That's a typical follow-up question.
(ii) $y = \dfrac{3x+2}{5x+6}$
$u = 3x+2$ $v = 5x+6$
$\dfrac{du}{dx} = 3$ $\dfrac{dv}{dx} = 5$
$\dfrac{dy}{dx} = \dfrac{3(5x+6) - 5(3x+2)}{(5x+6)^{2}}$
$\dfrac{dy}{dx} = \dfrac{15x + 18 - 15x - 10}{(5x+6)^{2}}$
$\dfrac{dy}{dx} = \dfrac{8}{(5x+6)^{2}}$
(iii) $y = \dfrac{7x-3}{5-x}$ (where $x$ ≠ 5)
Careful — bottom is $5-x$, so $\dfrac{dv}{dx} = -1$, not $+1$.
$u = 7x-3$ $v = 5-x$
$\dfrac{du}{dx} = 7$ $\dfrac{dv}{dx} = -1$
$\dfrac{dy}{dx} = \dfrac{7(5-x) - (7x-3)(-1)}{(5-x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{7(5-x) + (7x-3)}{(5-x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{35 - 7x + 7x - 3}{(5-x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{32}{(5-x)^{2}}$
Section 6 of 11
Quotient Rule — Your Turn
You try
Differentiate $\;y = \dfrac{2x+1}{3x+5}$.
$u = 2x+1$, $v = 3x+5$. Then $\dfrac{v \cdot du - u \cdot dv}{v^{2}}$.
$u = 2x+1$, $\;\;v = 3x+5$
$\dfrac{du}{dx} = 2$, $\;\;\dfrac{dv}{dx} = 3$
$\dfrac{dy}{dx} = \dfrac{2(3x+5) - 3(2x+1)}{(3x+5)^{2}}$
$= \dfrac{6x + 10 - 6x - 3}{(3x+5)^{2}}$
$\dfrac{dy}{dx} = \dfrac{7}{(3x+5)^{2}}$
$\dfrac{7}{(3x+5)^{2}}$
You try
Differentiate $\;y = \dfrac{4x-1}{2x+3}$.
$u = 4x-1$, $v = 2x+3$. Watch the minus carefully.
$u = 4x-1$, $\;\;v = 2x+3$
$\dfrac{du}{dx} = 4$, $\;\;\dfrac{dv}{dx} = 2$
$\dfrac{dy}{dx} = \dfrac{4(2x+3) - 2(4x-1)}{(2x+3)^{2}}$
$= \dfrac{8x + 12 - 8x + 2}{(2x+3)^{2}}$
$\dfrac{dy}{dx} = \dfrac{14}{(2x+3)^{2}}$
$\dfrac{14}{(2x+3)^{2}}$
You try
Differentiate $\;y = \dfrac{5x+2}{x-3}$.
$u = 5x+2$, $v = x-3$. Both $\dfrac{du}{dx}$ and $\dfrac{dv}{dx}$ are easy here.
$u = 5x+2$, $\;\;v = x-3$
$\dfrac{du}{dx} = 5$, $\;\;\dfrac{dv}{dx} = 1$
$\dfrac{dy}{dx} = \dfrac{5(x-3) - 1(5x+2)}{(x-3)^{2}}$
$= \dfrac{5x - 15 - 5x - 2}{(x-3)^{2}}$
$\dfrac{dy}{dx} = \dfrac{-17}{(x-3)^{2}}$
$-\dfrac{17}{(x-3)^{2}}$
Section 7 of 11
The Chain Rule
When you see a function of a function — something inside a bracket, raised to a power, or fed into $\sin$, $\cos$, $\ln$ or $e$ — that's the Chain Rule.
Think of it like a Russian doll. You differentiate the outside, leaving the inside alone, then multiply by the derivative of the inside.
Chain Rule — Must learn
$\dfrac{dy}{dx} = \dfrac{du}{dx} \cdot \dfrac{dy}{du}$
Let $u$ be the inside. Then $y$ becomes a function of $u$. Differentiate both pieces and multiply.
Worked example: $y = (3x+1)^{5}$
The inside is $3x+1$, raised to the 5. Let $u$ be the inside.
$u = 3x+1$ $\dfrac{du}{dx} = 3$
$y = u^{5}$ $\dfrac{dy}{du} = 5u^{4} = 5(3x+1)^{4}$
$\dfrac{dy}{dx} = \dfrac{du}{dx} \cdot \dfrac{dy}{du} = 3 \cdot 5(3x+1)^{4}$
$\dfrac{dy}{dx} = 15(3x+1)^{4}$
The shortcut (once you're confident)
Differ the inside × differ the outside, leaving the inside alone.
For $(3x+1)^{5}$: inside-diff is $3$, outside-diff is $5(3x+1)^{4}$. Multiply: $15(3x+1)^{4}$.
Section 8 of 11
Chain Rule — More Examples
(i) $y = (7x+1)^{6}$
Inside-diff: $7$ Outside-diff: $6(7x+1)^{5}$
$\dfrac{dy}{dx} = 7 \cdot 6(7x+1)^{5}$
$\dfrac{dy}{dx} = 42(7x+1)^{5}$
(ii) $y = (6x+3)^{10}$
Inside-diff: $6$ Outside-diff: $10(6x+3)^{9}$
$\dfrac{dy}{dx} = 6 \cdot 10(6x+3)^{9}$
$\dfrac{dy}{dx} = 60(6x+3)^{9}$
(iii) $y = \sin(3x+1)$
The "outside" is $\sin$. Its derivative is $\cos$, with the same inside left alone.
Inside-diff: $3$ Outside-diff: $\cos(3x+1)$
$\dfrac{dy}{dx} = 3 \cos(3x+1)$
(iv) $y = \sin(8x+11)$
Inside-diff: $8$ Outside-diff: $\cos(8x+11)$
$\dfrac{dy}{dx} = 8 \cos(8x+11)$
(v) $y = \cos(4x-1)$
Outside is $\cos$ — its derivative is $-\sin$. Don't forget the minus.
Inside-diff: $4$ Outside-diff: $-\sin(4x-1)$
$\dfrac{dy}{dx} = 4 \cdot (-\sin(4x-1))$
$\dfrac{dy}{dx} = -4 \sin(4x-1)$
Section 9 of 11
Chain Rule — Your Turn
You try
Differentiate $\;y = (5x+3)^{4}$.
Inside-diff $\times$ outside-diff. Outside-diff: bring down the 4, reduce power to 3.
Inside-diff: $5$, Outside-diff: $4(5x+3)^{3}$
$\dfrac{dy}{dx} = 5 \cdot 4(5x+3)^{3}$
$\dfrac{dy}{dx} = 20(5x+3)^{3}$
$20(5x+3)^{3}$
You try
Differentiate $\;y = (4x-1)^{7}$.
Inside-diff is $4$. Outside-diff: $7(4x-1)^{6}$.
$\dfrac{dy}{dx} = 4 \cdot 7(4x-1)^{6}$
$\dfrac{dy}{dx} = 28(4x-1)^{6}$
$28(4x-1)^{6}$
You try
Differentiate $\;y = \sin(7x)$.
Inside-diff is $7$. Sin becomes cos, inside left alone.
Inside-diff: $7$, Outside-diff: $\cos(7x)$
$\dfrac{dy}{dx} = 7 \cos(7x)$
$7 \cos(7x)$
You try
Differentiate $\;y = \cos(2x+5)$.
Inside-diff is $2$. Cos becomes $-\sin$. Watch the minus.
Inside-diff: $2$, Outside-diff: $-\sin(2x+5)$
$\dfrac{dy}{dx} = 2 \cdot (-\sin(2x+5))$
$\dfrac{dy}{dx} = -2 \sin(2x+5)$
$-2 \sin(2x+5)$
Section 10 of 11
Mix it Up — Identify the Rule
Before you can differentiate, you have to pick the right rule. Use the $x = 10$ trick if you're not sure.
By $\Rightarrow$ Product · Divide by $\Rightarrow$ Quotient · Of $\Rightarrow$ Chain · otherwise just the Rule (multiply by power, reduce by 1).
You try
Which rule for $\;y = (3x-1)\sin x$?
Two pieces. What joins them?
At $x = 10$: $(3(10)-1) = 29$ and $\sin 10 \approx -0.544$
$y = (29)(-0.544)$ — two answers multiplied
Product Rule
Product Rule
You try
Which rule for $\;y = \dfrac{x+1}{x^{2}+1}$?
Top and bottom — what joins them?
Top: $11$. Bottom: $101$. $y = \dfrac{11}{101}$.
Two answers divided.
Quotient Rule
Quotient Rule
You try
Which rule for $\;y = (9x+2)^{5}$?
A bracket raised to a power — the 5th power of the bracket.
At $x = 10$: bracket gives $92$, then we raise it to the 5.
One operation, "of" the bracket.
Chain Rule
Chain Rule
You try
Which rule for $\;y = 4x^{3} + 7x - 2$?
No brackets, no division, no "of". Just a sum of powers.
Sum of powers of $x$ — standard form.
Just the Rule from Part 1: multiply by power, reduce by 1.
The Rule (Part 1)
Section 11 of 11
Mix it Up — Differentiate
Now pick the rule and differentiate. Identify first, then go.
You try
Differentiate $\;y = (4x+3)^{6}$.
Bracket to a power — Chain. Inside-diff $\times$ outside-diff.
Chain Rule. Inside-diff: $4$, Outside-diff: $6(4x+3)^{5}$.
$\dfrac{dy}{dx} = 4 \cdot 6(4x+3)^{5}$
$\dfrac{dy}{dx} = 24(4x+3)^{5}$
$24(4x+3)^{5}$
You try
Differentiate $\;y = \dfrac{x+2}{x-2}$.
Divided — Quotient. $u = x+2$, $v = x-2$.
Quotient Rule. $u = x+2$, $\;v = x-2$.
$\dfrac{du}{dx} = 1$, $\;\dfrac{dv}{dx} = 1$.
$\dfrac{dy}{dx} = \dfrac{1(x-2) - 1(x+2)}{(x-2)^{2}} = \dfrac{x - 2 - x - 2}{(x-2)^{2}}$
$\dfrac{dy}{dx} = \dfrac{-4}{(x-2)^{2}}$
$-\dfrac{4}{(x-2)^{2}}$
You try
Differentiate $\;y = x^{2} \cos x$.
Multiplied — Product. $u = x^{2}$, $v = \cos x$. Remember $\dfrac{d}{dx}(\cos x) = -\sin x$.
Product Rule. $u = x^{2}$, $\;v = \cos x$.
$\dfrac{du}{dx} = 2x$, $\;\dfrac{dv}{dx} = -\sin x$.
$\dfrac{dy}{dx} = x^{2}(-\sin x) + \cos x \cdot 2x$
$\dfrac{dy}{dx} = -x^{2} \sin x + 2x \cos x$
$2x \cos x - x^{2} \sin x$
You try
Differentiate $\;y = \sin(5x+3)$.
Sin "of" something — Chain. Inside-diff $\times$ outside-diff.
Chain Rule. Inside-diff: $5$, Outside-diff: $\cos(5x+3)$.
$\dfrac{dy}{dx} = 5 \cos(5x+3)$
$5 \cos(5x+3)$
That's Differentiation — Part 2.
Product · Quotient · Chain. You can now spot which rule to use and apply each one cleanly. Coming next: Logs, Exponentials, and Inverse Trig — built on top of the Chain Rule.