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?

Jun 18, 2018

False

Explanation:

Take $f \left(x\right) = {x}^{2} + 1 , g \left(x\right) = x$
then

$f \left(x\right) g \left(x\right) = {x}^{3} + x$
then
$f ' \left(x\right) \cdot g ' \left(x\right) = 2 x$
But

$\left(f \left(x\right) g \left(x\right)\right) ' = 3 {x}^{2} + 1$
It must be
$\left(f \left(x\right) g \left(x\right)\right) ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

Jun 18, 2018

The statement is false .

The product rule provides the correct formulation:

$y = f \left(x\right) g \left(x\right) \implies y = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

Explanation:

We can readily disprove the given statement:

Consider:

$f \left(x\right) = x$ and $g \left(x\right) = x$

Then differentiating wrt $x$ we have:

$f ' \left(x\right) = 1$ and $g ' \left(x\right) = 1 \implies f ' \left(x\right) g ' \left(x\right) = 1$

And $y = f \left(x\right) g \left(x\right) = {x}^{2} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \ne f ' \left(x\right) g ' \left(x\right)$

And so By counterexample, the statement is false .

In fact the product rule provides the correct formulation:

$y = f \left(x\right) g \left(x\right) \implies y = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$