# Black-Scholes Implied Volatility

I'm working my way through the following paper:

Malz. A. M. (2014). A Simple and Reliable Way to Compute Option-Based Risk-Neutral Distributions

I am completely stuck on the following derivation. The author expresses the price of a call option at time-$$t$$ (with strike $$K$$ and on underlier $$S_t$$) as $$$$c(t; K, T) = v[S_t , K, T, σ_\text{imp}(t, K), r].$$$$

where $$v(.)$$ denotes the Black-scholes pricing formula for a European call, and $$\sigma_\text{imp}(t, K)$$ is the B-S implied vol.

I understand this, but the following step is not clear to me. The author differentiates both sides of the above equation with respect to the strike, $$K$$. This gives:

$$$$\frac{\partial c}{\partial K} = \frac{\partial v}{\partial K} + \frac{\partial v}{\partial \sigma_\text{imp}}\frac{\partial \sigma_\text{imp}}{\partial K}$$$$

But how can this be true?

• It is a slight modification of Black Scholes- it assumes implied volatility varies with Strike (which is what one sees in practice: smile). Then the derivative is just what is called total derivative: when k changes, option price changes, but the volatility also changes with k, which has an additional impact on the price. Pls see here for the definition: mathworld.wolfram.com/TotalDerivative.html – Magic is in the chain Jul 6 at 21:59
• precisely, the volatility varies with the strike hence you must perform a partial derivative there too. you can think of it as dc/dK = dv/dK (straight derivative) – John Jul 7 at 1:21

$$f(x) = g(h(x),k(x))$$
$$f'(x) = \partial_1 g (h(x),k(x)) h'(x) + \partial_2 g (h(x),k(x)) k'(x)$$
(ignore red herrings: $$t$$, $$T$$, $$S_t$$ and $$r$$; focus on $$K$$ only)