Derivative of $\sqrt{x - 1} (x + 1)$ $sqrt(x-1)(x+1)$ ?

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Answer 1 · 2017-11-20T03:42:36

$\frac{d y}{d x} = \frac{(3 x - 1) \sqrt{x - 1}}{2 (x - 1)}$ $dy/dx=((3x-1)sqrt(x-1))/(2(x-1))$

This is a product of two functions. We have to use the product rule:

$y = \sqrt{x - 1} (x + 1) = f (x) g (x)$ $y=sqrt(x-1)(x+1)=f(x)g(x)$

$f (x) = \sqrt{x - 1} = {(x - 1)}^{\frac{1}{2}} = {(x - 1)}^{\frac{1}{2}}$ $f(x)=sqrt(x-1)=(x-1)^(1/2)=(x-1)^(1/2)$

$g (x) = (x + 1)$ $g(x)=(x+1)$

$f'(x)=1/2(x-1)^(1/2-1)=1/2(x-1)^(-1/2)(1)=1/(2(x-1)^(1/2))=1/(2sqrt(x-1))$

$g'(x)=1$

$dy/dx=f'(x)g(x)+g'(x)f(x)$

$dy/dx=1/(2sqrt(x-1))(x+1)+(1)(sqrt(x-1))$

$dy/dx=(x+1)/(2sqrt(x-1))+sqrt(x-1)$

$dy/dx=((x+1)+2sqrt(x-1)sqrt(x-1))/(2sqrt(x-1))$

$dy/dx=(x+1+2(x-1))/(2sqrt(x-1)$

$dy/dx=(x+1+2x-2)/(2sqrt(x-1))=(3x-1)/(2sqrt(x-1)$

Now we can multiply both top and bottom by $sqrt(x-1)$ to normalize it:

$dy/dx=((3x-1)sqrt(x-1))/(2sqrt(x-1)sqrt(x-1))=((3x-1)sqrt(x-1))/(2(x-1))$

Derivative of sqrt(x-1)(x+1)√x−1(x+1)sqrt(x-1)(x+1)?