Derivatives of standard functions via chain rule

Before the chain rule can do any work, the standard derivatives must be automatic — they are the "atoms" that every composite decomposes into.

where: each rule gives the derivative of the named function of the single variable $x$ itself, with $a>0$, $a\neq1$ constant in the last case; these are used as building blocks, not re-derived here. Now the rule that combines them. If $y=f(u)$ and $u=g(x)$, so $y=f(g(x))$ is a composite of two layers, then

where: $u=g(x)$ is the inner function, $f(u)$ is the outer function, $f'(g(x))$ is the outer derivative evaluated at the inner function (inner left untouched), and $g'(x)$ is the inner derivative. For a chain of three or more layers the same idea repeats: differentiate the outermost layer first, then multiply successively by the derivative of each layer moving inward, one multiplied factor per layer — for $y=\ln(\sin(x^2))$ this gives three factors: the derivative of $\ln$ at $\sin(x^2)$, times the derivative of $\sin$ at $x^2$, times the derivative of $x^2$. The six inverse-trigonometric derivatives are obtained once, by substitution, and then reused as further standard building blocks:

where: each formula gives the derivative with respect to $x$ of the inverse trig function of $x$ itself; the derivation sets $y=\sin^{-1}x$ so $x=\sin y$, differentiates implicitly to get $1=\cos y\,\dfrac{dy}{dx}$, then rewrites $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$ (positive root, since $y$ lies in the principal-value range). Once memorised, the chain rule applies to them exactly as to $\sin x$ or $e^x$ whenever the argument is itself a function of $x$: $\dfrac{d}{dx}\sin^{-1}(3x)=\dfrac{3}{\sqrt{1-9x^2}}$, the standard result multiplied by the derivative of the inner function $3x$. An exam-level shortcut worth internalising: expressions like $\dfrac{2x}{1-x^2}$, $\dfrac{1-x^2}{1+x^2}$ or $3x-4x^3$ sitting inside an inverse-trig function often signal a substitution such as $x=\tan\theta$ or $x=\sin\theta$ that collapses the whole expression to a simple multiple of $\theta$ before any differentiation happens, turning a messy quotient-rule-inside-chain-rule problem into a one-line derivative — demonstrated in the derivation and second worked example below. It also helps to notice which standard derivative each layer is drawn from, since exam questions mix families freely: a single composite may combine a trigonometric outer layer with a polynomial inner layer, or an exponential outer layer with a logarithmic inner layer, or an inverse-trig outer layer with a rational inner layer. In every case the identification step — write down explicitly what $u$ is and what $f$ is applied to it — is what prevents errors; students who skip straight to differentiating "by feel" are the ones who most often forget the inner-function factor.

Differentiate $y = \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$ with respect to $x$, for $-1<x<1$.

Step 1. Recognise the standard substitution. The form $\dfrac{2x}{1-x^2}$ matches the double-angle identity $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$, so try $x=\tan\theta$.

Step 2. Substitute. Let $x=\tan\theta$, so $\theta=\tan^{-1}x$. Then

Step 3. Differentiate the simplified form using the standard inverse-trig derivative and the chain rule (here just a constant multiple, since the inner function is $x$ itself):

Step 4. State the final answer.

This is far faster than attempting the quotient rule inside an inverse-trig chain rule directly — spotting the compound-angle pattern first is a key exam skill.

Competitive stretch

For the triple composition $y=\ln\big(\sin(x^2)\big)$, the same layer-peeling gives three multiplied factors: $\dfrac{dy}{dx}=\dfrac{1}{\sin(x^2)}\cdot\cos(x^2)\cdot 2x=2x\cot(x^2)$ — outer $\ln$, middle $\sin$, inner $x^2$, each contributing exactly one factor.

• Forgetting to multiply by the derivative of the inner function — writing $\frac{dy}{dx}=\cos(3x^2+2x)$ and stopping is the single most common chain-rule error.
• Misapplying inverse-trig derivative signs, especially confusing the negative signs of $\cos^{-1}x$ and $\sec^{-1}x$ with the positive signs of $\sin^{-1}x$ and $\tan^{-1}x$.
• Not restricting the domain when using a trigonometric substitution such as $x=\tan\theta$ — the simplification $\tan^{-1}(\tan2\theta)=2\theta$ only holds within the correct principal-value range.
• Applying the chain rule only once for a doubly- or triply-nested function instead of peeling off every layer in turn.
• Forgetting that $a^x$ differentiates to $a^x\ln a$, not $a^x$ alone and not $xa^{x-1}$ — a distinct rule easily blurred with the power rule or the $e^x$ rule.
• Using a trigonometric substitution outside the range where the corresponding compound-angle identity actually holds, which can silently flip a sign in the simplified derivative without any warning in the algebra.
• Trying to differentiate a quotient like $\dfrac{2x}{1-x^2}$ directly with the quotient rule inside an inverse-trig chain rule instead of first checking whether it matches a known compound-angle pattern — the direct route is correct but far slower and much more error-prone under exam time pressure.

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