 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/15/b/i/solution

Assume the needed derivatives exist continuously and

df / d g \neq = 0

on a neighborhood, so the inverse function theorem provides a local inverse. Put

p = df / d g

q = d^{2} f / d g^{2}

, and

r = d^{3} f / d g^{3}

. The chain rule gives

d / df = p^{- 1} d / d g

, so

\frac{d g}{df} = \frac{1}{p}, \frac{d ^{2} g}{d f ^{2}} = \frac{1}{p} \frac{d}{d g} \frac{1}{p} = - \frac{q}{p ^{3}} .

Differentiate once more:

\frac{d ^{3} g}{d f ^{3}} = \frac{1}{p} \frac{d}{d g} (- q p^{- 3}) = - \frac{r}{p ^{4}} + \frac{3 q ^{2}}{p ^{5}} .

\frac{d g}{df} = \frac{1}{df / d g}, \frac{d ^{2} g}{d f ^{2}} = - \frac{d ^{2} f / d g ^{2}}{( df / d g ) ^{3}}, \frac{d ^{3} g}{d f ^{3}} = \frac{3 ( d ^{2} f / d g ^{2} ) ^{2} - ( df / d g ) ( d ^{3} f / d g ^{3} )}{( df / d g ) ^{5}} .

All forward derivatives on the right are evaluated at the corresponding value

g (f)

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