Imagine you're assembling a toy. First you put part A into part B, then you put that combined piece into part C. The final toy's position depends on how you moved A, which then affected B, which then affected C. That's exactly what the chain rule captures — how a change in the first variable ripples through a sequence of functions to affect the final output.
Let's make this concrete. Suppose you have a function f that depends on g, and g itself depends on x:
y=f(g(x))
You want to know: if x changes by a tiny amount, how much does y change? The answer isn't just f′(g(x)) — because g(x) itself changes when x changes. You have to multiply the two rates:
How fast does g change with respect to x? That's g′(x).
How fast does f change with respect to its input g? That's f′(g(x)).
The total effect is the product:
dxdy=f′(g(x))⋅g′(x)
Note
In Leibniz notation, this looks even more natural: dxdy=dudy⋅dxdu, where u=g(x). The du's "cancel" like fractions — though this is just a helpful memory aid, not a rigorous proof.
The Precise Statement
Chain Rule (single variable): If g is differentiable at x and f is differentiable at g(x), then the composite function h(x)=f(g(x)) is differentiable at x, and
h′(x)=f′(g(x))⋅g′(x)
That's it. One multiplication. But the power is enormous — it lets you differentiate almost any nested function.
A Simple Example
Differentiate h(x)=sin(3x2).
Here f(u)=sinu and g(x)=3x2. Then:
f′(u)=cosu, so f′(g(x))=cos(3x2)
g′(x)=6x
Multiply: h′(x)=cos(3x2)⋅6x=6xcos(3x2)
Watch out
The most common mistake is forgetting to multiply by the inner derivative. Students often write dxdsin(3x2)=cos(3x2) and stop — that's wrong. The chain rule demands you also multiply by 6x.
Why It's Called a "Chain"
Think of a chain of links: x→g→f. Each link has its own rate of change. To find the total rate from x to f, you multiply the rates of each link. If you had three functions — say h(x)=f(g(k(x))) — you'd multiply three derivatives: …
Substituting x=sinθ simplifies y to 2sin−1x for −21<x<21; differentiating gives dxdy=1−x22.
Recognising the Identity
The expression 2x1−x2 is the double-angle formula sin2θ=2sinθcosθ in terms of x=sinθ: once 1−x2 is identified as cosθ, the whole argument collapses to sin2θ.
Step-by-Step Solution
1. Substitute x=sinθ.
Let θ=sin−1x∈(−2π,2π). Since −21<x<21, we get θ∈(−4π,4π), so cosθ>0 throughout, and:
1−x2=1−sin2θ=cos2θ=cosθ.
Hence:
2x1−x2=2sinθcosθ=sin2θ⇒y=sin−1(sin2θ).
2. Check the range of 2θ.
Since θ∈(−4π,4π):
2θ∈(−2π,2π),
exactly the principal range of sin−1. So sin−1(sin2θ)=2θ directly, with no correction needed.
Method: Simplify Inverse-Trig Compositions by Trigonometric Substitution
This method applies whenever you must differentiate an inverse trig function whose argument is built from x and 1−x2 using a double-angle-style pattern — direct differentiation of the raw expression would be painful, so simplify first.
Steps
Step 1: Spot the double-angle pattern and substitute
Look at the argument inside the inverse trig function. An expression like 2x1−x2 matches 2sinθcosθ=sin2θ. Substitute x=sinθ, restricting θ to the principal branch of the inverse function you substituted with.
Step 2: Rewrite the argument using the identity
1−x2=1−sin2θ=cosθ(positive on the given domain)
so the whole argument collapses to a single trig function of 2θ.
Mistake 1: Stopping at the simplified expression for y instead of differentiating
The substitution trick reduces y to something clean like 2sin−1x — but the question asks for dxdy, not for a simplified y. Why it's wrong: simplifying is only the first half of the problem. Correct approach: always take the extra step and differentiate the simplified form, e.g. dxd(2sin−1x)=1−x22.
Mistake 2: Ignoring the domain restriction when reducing sin−1(sin2θ) to 2θ …