# What is the second derivative of (f * g)(x) if f and g are functions such that f'(x)=g(x) and g'(x)=f(x)?

Mar 19, 2018

$\left(4 f \cdot g\right) \left(x\right)$

#### Explanation:

Let $P \left(x\right) = \left(f \cdot g\right) \left(x\right) = f \left(x\right) g \left(x\right)$

Then using the product rule:

$P ' \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.

Using the condition given in the question, we get:

$P ' \left(x\right) = {\left(g \left(x\right)\right)}^{2} + {\left(f \left(x\right)\right)}^{2}$

Now using the power and chain rules:

$P ' ' \left(x\right) = 2 g \left(x\right) g ' \left(x\right) + 2 f \left(x\right) f ' \left(x\right)$.

Applying the special condition of this question again, we write:

$P ' ' \left(x\right) = 2 g \left(x\right) f \left(x\right) + 2 f \left(x\right) g \left(x\right) = 4 f \left(x\right) g \left(x\right) = 4 \left(f \cdot g\right) \left(x\right)$

Mar 19, 2018

Another answer in case $f \cdot g$ is meant to be the composition of $f$ and $g$

#### Explanation:

We want to find the second derivative of $\left(f \cdot g\right) \left(x\right) = f \left(g \left(x\right)\right)$

We differentiate once using the chain rule.

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right) = f ' \left(g \left(x\right)\right) f \left(x\right)$

Then we differentiate again using the product chain rules

$\frac{d}{\mathrm{dx}} f ' \left(g \left(x\right)\right) f \left(x\right) = f ' ' \left(g \left(x\right)\right) g ' \left(x\right) f \left(x\right) + f ' \left(x\right) f ' \left(g \left(x\right)\right)$

$= f ' ' \left(g \left(x\right)\right) {\left[f \left(x\right)\right]}^{2} + g \left(x\right) f ' \left(g \left(x\right)\right)$