Filling the quotient template
With the blocks expanded, insert them into the skeleton:
\[h'(x)=\frac{\bige^{2x}\cdot 2\big-e^{2x}\big[-2x\big]}{\big(1-x^2\big)^2}\]
Each block sits in its own slot. The numerator factor \(e^{2x}\) is untouched in the second term; the denominator factor \(1-x^2\) is untouched in the first term. The denominator of the result is the whole original denominator squared.
The factor audit
Before simplifying, run three checks.
Check 1: one chain rule block per composite part
The numerator \(e^{2x}\) is a composite, so it contributes one block. The denominator \(1-x^2\) is a composite, so it contributes one block. Two composite parts, two blocks. If you count only one block, you have dropped a layer.
Check 2: the denominator is squared as a whole
The result's denominator is \((1-x^2)^2\), not \(-2x\) and not \(1-x^2\). Squaring the derivative of the denominator is a common error; the quotient rule always squares the original denominator.
Check 3: the inner expression is preserved inside each outer factor
In the first block, \(e^{2x}\) still carries the exponent \(2x\), not \(x\). In the second block, the derivative \(-2x\) is the derivative of the whole inner expression \(x^2\), not of \(x\) alone.
Why the audit works
Each check targets a different failure mode: a dropped layer, a mis-squared denominator, and a drifted argument. Running all three before simplifying catches the error while the structure is still visible. Simplifying first hides which factor came from which layer.
What this page establishes
The quotient rule derivative is complete with both chain rule blocks in place, and the three-check audit confirms no factor was dropped. The same audit applies to any product or quotient of composites.