How do you find the derivative of $y = ln (x + 3) ln (x - 1)$ $y=ln(x+3)ln(x-1)$ ?

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Answer 1 · 2017-01-13T13:17:06

$\frac{d}{d x} (ln (x + 3) ln (x + 1)) = \frac{ln (x + 1)}{x + 3} + \frac{ln (x + 3)}{x + 1}$ $d/(dx)(ln(x+3)ln(x+1)) = ln(x+1)/(x+3)+ln(x+3)/(x+1)$

We can use the product rule:

$d/(dx)( f(x)*g(x)) = f'(x) g(x) +f(x) g'(x)$

So:

$d/(dx)(ln(x+3)ln(x+1)) = ln(x+1)/(x+3)+ln(x+3)/(x+1)$

How do you find the derivative of y=ln(x+3)ln(x-1)y=ln(x+3)ln(x−1)y=ln(x+3)ln(x-1)?