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Neporo4naja [7]
3 years ago
8

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

Mathematics
1 answer:
Nadya [2.5K]3 years ago
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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