Question

f (x) = x² + 1 / x - 3 the first derivative is​

Answer 1

Answer :-

= x² - 6x - 1 / x² - 6x + 9

Explanation :-

f(x) = x² + 1 / x - 3

u = x² + 1

v = x - 3

u' = 2x

v' = 1

f' (x) = u' . v - v' . u / v²

       = 2x . (x - 3) - 1 . x² + 1 / (x - 3)²

       = 2x² - 6x - x² - 1 / (x - 3)²

       = x² - 6x - 1 / x² - 6x + 9

Answer 2

Answer:

$\dfrac{d}{dx}\left(\frac{x^2+1}{x-3}\right)=\dfrac{x^2-6x-1}{\left(x-3\right)^2}$

Step-by-step explanation:

$\dfrac{d}{dx}\left(\dfrac{x^2+1}{x-3}\right)$

$\mathrm{Apply\:the\:Quotient\:Rule}:\quad \left(\dfrac{f}{g}\right)^'=\dfrac{f\:'\cdot g-g'\cdot f}{g^2}$

$=\dfrac{\dfrac{d}{dx}\left(x^2+1\right)\left(x-3\right)-\dfrac{d}{dx}\left(x-3\right)\left(x^2+1\right)}{\left(x-3\right)^2}$

$\dfrac{d}{dx}\left(x^2+1\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\dfrac{d}{dx}\left(x^2\right)+\dfrac{d}{dx}\left(1\right)$

$=2x+0$

$\mathrm{Simplify}$

$=2x$

$\dfrac{d}{dx}\left(x-3\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\dfrac{d}{dx}\left(x\right)-\dfrac{d}{dx}\left(3\right)$

$=1-0$

$\mathrm{Simplify}$

$=1$

$=\dfrac{2x\left(x-3\right)-1\cdot \left(x^2+1\right)}{\left(x-3\right)^2}$

$\mathrm{Simplify\:}\dfrac{2x\left(x-3\right)-1\cdot \left(x^2+1\right)}{\left(x-3\right)^2}$

$\mathrm{Multiply:}\:1\cdot \left(x^2+1\right)=\left(x^2+1\right)$

$=\dfrac{2x\left(x-3\right)-\left(x^2+1\right)}{\left(x-3\right)^2}$

$\mathrm{Expand}\:2x\left(x-3\right)-\left(x^2+1\right)$

$\mathrm{Expand}\:2x\left(x-3\right):\quad 2x^2-6x$

$=2x^2-6x-\left(x^2+1\right)$

$-\left(x^2+1\right):\quad -x^2-1$

$-\left(x^2+1\right):\quad -x^2-1$

$\mathrm{Simplify}\:2x^2-6x-x^2-1:\quad x^2-6x-1$

$=x^2-6x-1$

$=\dfrac{x^2-6x-1}{\left(x-3\right)^2}$