How do you find the derivative of $\frac{x - 3}{2 x + 1}$ $(x-3)/(2x+1)$ ?

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How do you find the derivative of $\frac{x - 3}{2 x + 1}$ $(x-3)/(2x+1)$ ?

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Answer 1 · 2017-10-05T20:14:52

$\frac{7}{{(2 x + 1)}^{2}}$ $7/(2x+1)^2$

Explanation:

$differentiate using the quotient rule$ $"differentiate using the "color(blue)"quotient rule"$

$given f (x) = \frac{g (x)}{h (x)} then$ $"given "f(x)=(g(x))/(h(x))" then"$

$f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"$

$g(x)=x-3rArrg'(x)=1$

$h(x)=2x+1rArrh'(x)=2$

$rArrf'(x)=((2x+1)-2(x-3))/(2x+1)^2$

$color(white)(rArrf'(x))=7/(2x+1)^2$

Answer 2 · 2017-10-05T20:19:51

$=7/(4x^2-4x+1)$ or $7/(2x+1)^2$

Explanation:

Quotient rule: $(u'v )-(uv')$ all divided by $v^2$ , assuming the top line is $u$ and bottom is $v$

In this function, $u= x-3$ and $v=2x+1$

$:. u' = 1 " and " v' = 2$

So plugging that in gets:

$(1*(2x+1)-(x-3)*2)/(2x+1)^2$

$=((2x+1)-(2x-6))/(4x^2+4x+1)$

$= (cancel(2x)+1-cancel(2x)+6)/(4x^2-4x+1)$

$=7/(4x^2-4x+1)$

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How do you find the derivative of $\frac{x - 3}{2 x + 1}$ $(x-3)/(2x+1)$ ?

2 Answers

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How do you find the derivative of $\frac{x - 3}{2 x + 1}$ $(x-3)/(2x+1)$ ?