Delta of a European Call: Why the Extra Terms Cancel
Under the Black-Scholes-Merton assumptions, the price of a European call on a non-dividend-paying stock is $$c = S\,N(d_1) - K e^{-r\tau} N(d_2), \qquad d_1 = \frac{\ln(S/K) + (r + \sigma^2/2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2 = d_1 - \sigma\sqrt{\tau},$$ where $\tau = T - t$ is the time to maturity and $N$ is the standard normal CDF.
What is the delta $\partial c/\partial S$ of this call? Derive it carefully. A tempting shortcut treats $N(d_1)$ and $N(d_2)$ as constants and reads off $\Delta = N(d_1)$ directly; explain why the shortcut is not a valid derivation even though it lands on the right number, and show exactly what happens to the terms it ignores.
Hints
- Both $d_1$ and $d_2$ depend on $S$ through $\ln(S/K)$, so the product rule produces two extra terms: $S\,N'(d_1)\,\partial d_1/\partial S$ and $-Ke^{-r\tau}N'(d_2)\,\partial d_2/\partial S$.
- Note that $\partial d_1/\partial S = \partial d_2/\partial S = 1/(S\sigma\sqrt{\tau})$, because $d_1 - d_2 = \sigma\sqrt{\tau}$ does not involve $S$.
- Show the identity $S\,N'(d_1) = Ke^{-r\tau}N'(d_2)$ by writing $N'(d_2) = N'(d_1)\exp(d_1\sigma\sqrt{\tau} - \sigma^2\tau/2)$ and simplifying the exponent to $\ln(S/K) + r\tau$.
Worked Solution
How to Think About It: $c$ depends on $S$ in three places: the explicit factor $S$ in front of $N(d_1)$, and the arguments $d_1$ and $d_2$, which both contain $\ln S$. An honest derivative has three terms. The shortcut keeps only the first. It happens to be right because the other two are equal and opposite, and proving that is the whole exercise.
Quick Estimate: Numerically, with $S = K = 100$, $r = 5\%$, $\sigma = 20\%$, $\tau = 1$: $d_1 = 0.35$, so $N(d_1) = 0.637$. A finite-difference delta $[c(100.01) - c(99.99)]/0.02$ gives $0.637$ as well, so the cancellation is real, not approximate.
Formal Solution:
*Step 1 -- Full derivative by the chain rule.* $$\frac{\partial c}{\partial S} = N(d_1) + S\,N'(d_1)\frac{\partial d_1}{\partial S} - Ke^{-r\tau}N'(d_2)\frac{\partial d_2}{\partial S}.$$ Since $d_2 = d_1 - \sigma\sqrt{\tau}$ and the difference is independent of $S$, $$\frac{\partial d_1}{\partial S} = \frac{\partial d_2}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}}.$$ Hence $$\Delta = N(d_1) + \frac{1}{S\sigma\sqrt{\tau}}\left[S\,N'(d_1) - Ke^{-r\tau}N'(d_2)\right].$$
*Step 2 -- The key identity.* With $N'(x) = e^{-x^2/2}/\sqrt{2\pi}$, $$N'(d_2) = N'(d_1)\exp\!\left(\frac{d_1^2 - d_2^2}{2}\right) = N'(d_1)\exp\!\left(\frac{(d_1 - d_2)(d_1 + d_2)}{2}\right).$$ Now $d_1 - d_2 = \sigma\sqrt{\tau}$ and $d_1 + d_2 = \dfrac{2\ln(S/K) + 2r\tau}{\sigma\sqrt{\tau}}$, so the exponent is $\ln(S/K) + r\tau$ and $$N'(d_2) = N'(d_1)\,\frac{S}{K}e^{r\tau} \quad\Longrightarrow\quad Ke^{-r\tau}N'(d_2) = S\,N'(d_1).$$
*Step 3 -- Cancellation.* The bracket in Step 1 is identically zero, so $$\Delta = \frac{\partial c}{\partial S} = N(d_1).$$
*Step 4 -- Why the shortcut is not a derivation.* Treating $N(d_1)$ and $N(d_2)$ as constants ignores two terms that are individually nonzero (each equals $N'(d_1)/(\sigma\sqrt{\tau})$, about $1.9$ in the numerical example above). The result is correct only because of the identity in Step 2. The same identity is what makes the delta of a European put $N(d_1) - 1$ and the vega $S\sqrt{\tau}N'(d_1)$ come out clean; and for a digital option, whose price is $e^{-r\tau}N(d_2)$ alone, there is no cancellation and the delta is $e^{-r\tau}N'(d_2)/(S\sigma\sqrt{\tau})$, so the shortcut would give the wrong answer there.
Answer: $\Delta = N(d_1)$. The full derivative contains the extra terms $S\,N'(d_1)\,\partial d_1/\partial S - Ke^{-r\tau}N'(d_2)\,\partial d_2/\partial S$, which cancel exactly because $\partial d_1/\partial S = \partial d_2/\partial S$ and $S\,N'(d_1) = Ke^{-r\tau}N'(d_2)$.
Intuition
The call formula reads like a portfolio of $N(d_1)$ shares and $-N(d_2)$ discounted strikes, and delta really is $N(d_1)$, but only because of a hidden identity: $S\,N'(d_1) = Ke^{-r\tau}N'(d_2)$. Both extra terms from the chain rule are the same number with opposite signs, so they cancel exactly. This is the same identity that keeps the replicating portfolio self-financing, and it is why the "shares plus bonds" reading of Black-Scholes is a genuine hedge rather than a coincidence. Interviewers use this question to see whether you differentiate honestly or just pattern-match the formula.