# Expressing the local volatility in terms of the implied volatility

Using the integral form of a call option price, I can derive the Dupire equation (no dividend) for the local volatility:

$$\sigma(S,T)^2=\frac{\frac{\partial C}{\partial T}+rK\frac{\partial C}{\partial K}}{\frac{1}{2}K^2\frac{\partial^2 C}{\partial K^2}}$$

However I have seen that this can be expressed in terms of the implied volatility, when assuming the Black-Scholes Model as:

$$\sigma(S,T)^2=\frac{\hat{\sigma}^2+2\hat{\sigma}(T-t)\frac{\partial \hat{\sigma}}{\partial T}+2r\hat{\sigma}K(T-t)\frac{\partial \hat{\sigma}}{\partial K}}{\left (1+Kd_1\sqrt{T-t}\frac{\partial \hat{\sigma}}{\partial K}\right )^2+\hat{\sigma}(T-t)K^2\left ( \frac{\partial^2\hat{\sigma}}{\partial K^2}-d_1\left ( \frac{\partial \hat{\sigma}}{\partial K}\right )^2\sqrt{T-t}\right )}$$

Now I've tried to get to this by using the chain rule: $$\frac{\partial C}{\partial T}=\frac{\partial C}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial T}$$ and similarly for $K$. However I cannot seem to get an expression that's quadratic in $\hat{\sigma}$ for the numerator. Am I on the right lines?

You are missing terms. The partial derivatives in the first formula must take into account the explicit dependence of the option price on expiry $T$ and strike $K$ and implicit dependence when using an implied volatility $\hat{\sigma}(T,K)$ that also depends on $T$ and $K$.
$$\frac{d}{dT} C(S,K,T, r, \hat{\sigma}(T,K)) = \frac{\partial C}{\partial T} + \frac{\partial C}{\partial \hat{\sigma}} \frac{\partial \hat{\sigma}}{\partial T}$$
• Thanks for your answer, to clarify, in the second derivative would we also have to take into account that the partial derivative $\frac{\partial C}{\partial T}$ also has a dependence on the implied vol. i.e.$$\frac{d^2 C}{dK^2}=\frac{\partial}{\partial K}\left ( \frac{\partial C}{\partial K}+\frac{\partial C}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial K}\right ) +\frac{\partial}{\partial \hat{\sigma}}\left (\frac{\partial C}{\partial K}+\frac{\partial C}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial K}\right )\frac{\partial \hat{\sigma}}{\partial K}$$ – George1811 Apr 7 '17 at 10:35