If f(x) = f(x)g(x), where f and g have derivatives of all orders, show that f'' = f ''g + 2f 'g' + fg''.

1 answer:

5 0

Differentiating once, we have

$f'(x)=f'(x)g(x)+f(x)g'(x)$

Differentiating again,

$f''(x)=f''(x)g(x)+f'(x)g'(x)+f'(x)g'(x)+f(x)g''(x)$
$f''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$

as needed.

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