First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 67
Let’s compute the derivatives on the right hand side of (2.9).
dφ(d 1 )
dσ
⎛
= d ( ∫ 1 d1
)
√ e −t2 /2 dt = 1
√ ⎜
d
dσ 2π −∞
2π
⎝dσ
∫ d1
−∞
e −t2 /2 dt⎟
⎠ .
} {{ }
u
∫ d1
We will use the chain rule to compute the derivative d e −t2 /2 dt
dσ
−∞
} {{ }
u
du
dσ = du dd 1
dd 1 dσ .
The first derivative follows from the Fundamental Theorem of Calculus
du
dd 1
= e −d 1 2 /2 ,
= du
dσ :
and the second derivative is an application of the quotient rule of differentiation
dd 1
dσ = d ( )
log(S/K)+(r + σ 2 /2)T
dσ
σ √ = √ T − log(S/K)+(r + σ2 /2)T
T
σ 2√ .
T
Putting the pieces together, we have
dφ(d 1 )
dσ
= e−d 1 2 /2
√
2π
( √T
−
log(S/K)+(r + σ 2 /2)T
σ 2√ T
Going back to the second derivative we need to compute in equation (2.9), we have:
dφ(d 2 )
dσ
= √ 1 ( ∫ d d2
)
e −t2 /2 dt .
2π dσ −∞
Using the chain rule and the Fundamental Theorem of Calculus we obtain
)
.
⎞
dφ(d 2 )
dσ
= e−d 2 2 /2
√
2π
dd 2
dσ .
Since d 2
is defined as d 2 = d 1 − σ √ T , we can express dd 2 /dσ in terms of dd 1 /dσ as:
dd 2
dσ = dd 1
dσ − √ T.