What is the derivative of $f(x) = (x − y)^2 g(x)$

Question

I'm studying about Newtons Method and my notes say that if

$f(x) = (x − y)^2 g(x)$ then $$f'(x) = 2(x-y)g(x) + (x − y)^2 g'(x)$$

Is this true? I thought it would be $$f'(x) = 2(x − y) g(x)$$

because $g(x)$ is just part of the multiplication.

Answer 1

If $$f(x)=(x-y)^2g(x)$$then $$f'(x)=2(x-y)g(x)+(x-y)^2g'(x)$$

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This happens because derivative of the product of two functions is the derivative of the first one multiplied by the second one plus the first one multiplied by the derivative of the second one.

i.e., $$(f\cdot g)'=f'\cdot g + f\cdot g'$$ and in your case $~y~$ is a constant.