If y = f(x)g(x), then dy/dx = f‘(x)g‘(x). If it is true, explain your answer. If false, provide a counterexample. True or False?

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Answer 1 · 2018-06-18T09:32:07

False

Take $f(x)=x^2+1,g(x)=x$
then

$f(x)g(x)=x^3+x$
then
$f'(x)*g'(x)=2x$
But

$(f(x)g(x))'=3x^2+1$
It must be
$(f(x)g(x))'=f'(x)*g(x)+f(x)*g'(x)$

Answer 2 · 2018-06-18T12:31:03

The statement is false .

The product rule provides the correct formulation:

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

We can readily disprove the given statement:

Consider:

$f(x)=x$ and $g(x)=x$

Then differentiating wrt $x$ we have:

$f'(x)=1$ and $g'(x)=1 => f'(x)g'(x)=1$

And $y=f(x)g(x) =x^2 => dy/dx = 2x != f'(x)g'(x)$

And so By counterexample, the statement is false .

In fact the product rule provides the correct formulation:

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