Differentiating quotients of functions using the quotient rule.
Answer at least 3 of 3 correctly to complete this section.
You have the product rule for f(x)×g(x). But what about g(x)f(x)?
Fractions arise constantly in calculus — from rational functions like x+1x2 to trigonometric ratios like cosxsinx=tanx. The quotient rule gives you a systematic way to differentiate any function divided by another.
There is a strong connection to the product rule: you can always rewrite vu as u⋅v−1 and use the product rule with the chain rule. The quotient rule is simply a shortcut that combines those steps. Master both approaches so you can pick the quickest method in an exam.
If y=vu, where u and v are both functions of x, then:
dxdy=v2vdxdu−udxdv
Or in function notation:
dxd[g(x)f(x)]=[g(x)]2f′(x)g(x)−f(x)g′(x)
Memory aid: “Low d-high minus high d-low, over the square of what’s below.” Here “high” is the numerator u and “low” is the denominator v.
The product rule gives u′v+uv′ — a plus. The quotient rule gives v2vu′−uv′ — a minus. This minus sign is the single most common source of errors. Be especially careful with it.
Write vu=u⋅v−1 and apply the product rule:
dxd[u⋅v−1]=u⋅(−v−2)dxdv+v−1⋅dxdu
=v1dxdu−v2udxdv=v2vdxdu−udxdv
This derivation shows the quotient rule is just the product rule in disguise.
Differentiate y=x+1x2
Step 1: u=x2, v=x+1.
Step 2: dxdu=2x, dxdv=1.
Step 3:
dxdy=(x+1)2(x+1)(2x)−x2(1)
Step 4: Simplify the numerator:
=(x+1)22x2+2x−x2=(x+1)2x2+2x=(x+1)2x(x+2)
Prove that dxd(tanx)=sec2x
Write tanx=cosxsinx.
Let u=sinx, v=cosx.
dxdu=cosx,dxdv=−sinx
dxd(tanx)=cos2xcosx⋅cosx−sinx⋅(−sinx)
=cos2xcos2x+sin2x=cos2x1=sec2x
This is a standard result you should memorise: dxd(tanx)=sec2x.
Answer at least 3 of 4 correctly to complete this section.
Before reaching for the quotient rule, always ask: can I simplify this fraction first?
If the denominator is a single term like x, x2, or x, you can divide each term in the numerator separately.
Differentiate y=xx3+4x
Simplify: y=x2+4
Differentiate: dxdy=2x
This is far quicker than the quotient rule. The quotient rule would give the same answer but with much more algebra.
Differentiate y=x23
Rewrite: y=3x−2
Differentiate: dxdy=−6x−3=−x36
No quotient rule needed — the power rule handles it.
The quotient rule is essential when the denominator is a sum or difference of terms and you cannot cancel:
| Expression | Simplify first? | Quotient rule? |
|---|---|---|
| xx3+2x | Yes — divide to get x2+2 | Not needed |
| x23 | Yes — rewrite as 3x−2 | Not needed |
| x+1x2 | No — cannot cancel | Yes |
| x2+1sinx | No — different function types | Yes |
| xex | No — cannot separate | Yes |
You always have a choice. For vu, you can either:
Both give the same answer. The product rule approach can be cleaner when the denominator is simple.
Exam tip: Use whichever method you are more confident with. Examiners accept both approaches for full marks.
Answer at least 3 of 4 correctly to complete this section.
The minus sign in the quotient rule is the single biggest source of lost marks. This section exists purely to make sure you never get it wrong.
dxdy=v2vdxdu−udxdv
The numerator is vu′−uv′, not uv′−vu′. Getting these the wrong way round flips the sign of your answer.
Answer at least 3 of 3 correctly to complete this section.
Differentiate y=x2sinx
Let u=sinx, v=x2.
dxdu=cosx,dxdv=2x
dxdy=(x2)2x2cosx−sinx⋅2x=x4x2cosx−2xsinx
Factorise x from the numerator:
dxdy=x4x(xcosx−2sinx)=x3xcosx−2sinx
Differentiate y=cosxex
Let u=ex, v=cosx.
dxdu=ex,dxdv=−sinx
dxdy=cos2xcosx⋅ex−ex⋅(−sinx)=cos2xex(cosx+sinx)
Note: cosxex=exsecx, so this is the same as saying dxd(exsecx)=cos2xex(cosx+sinx).
Differentiate y=x2+1e3x
Let u=e3x, v=x2+1.
dxdu=3e3x(chain rule),dxdv=2x
dxdy=(x2+1)2(x2+1)(3e3x)−e3x(2x)=(x2+1)2e3x(3x2+3−2x)
Answer at least 2 of 2 correctly to complete this section.
You always have a choice: use the quotient rule, or rewrite as a product with a negative power. This section shows the product rule approach so you can compare.
Differentiate y=e2xx3 using the product rule approach.
Rewrite: y=x3⋅e−2x
Let u=x3, v=e−2x.
dxdu=3x2,dxdv=−2e−2x
dxdy=x3(−2e−2x)+e−2x(3x2)=e−2x(3x2−2x3)=x2e−2x(3−2x)
This is often neater than the quotient rule for fractions with exponential denominators.
Answer at least 2 of 2 correctly to complete this section.
The curve C has equation y=x2+12x.
(a) Find dxdy.
(b) Find the coordinates of the stationary points.
(c) Determine their nature.
(a) Let u=2x, v=x2+1.
dxdu=2,dxdv=2x
dxdy=(x2+1)2(x2+1)(2)−2x(2x)=(x2+1)22x2+2−4x2=(x2+1)22−2x2=(x2+1)22(1−x2)
(b) Set dxdy=0:
The denominator (x2+1)2>0 for all x, so we need the numerator to be zero:
2(1−x2)=0⟹x2=1⟹x=±1
At x=1: y=22=1. Point: (1,1).
At x=−1: y=2−2=−1. Point: (−1,−1).
(c) We can use a sign-change test on dxdy. The sign of dxdy depends on 1−x2:
For x=1: 1−x2 changes from positive (when ∣x∣<1) to negative (when x>1), so gradient goes +→0→− → local maximum at (1,1).
For x=−1: 1−x2 changes from negative (when x<−1) to positive (when −1<x<1), so gradient goes −→0→+ → local minimum at (−1,−1).
Show that dxd(cotx)=−cosec2x
Write cotx=sinxcosx.
Let u=cosx, v=sinx.
dxdu=−sinx,dxdv=cosx
dxd(cotx)=sin2xsinx(−sinx)−cosx⋅cosx=sin2x−sin2x−cos2x=sin2x−1=−cosec2x
Answer at least 3 of 4 correctly to complete this section.
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