Question

d/dx(x^3+x^2+1)/(x^4+2x+1)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$<b>$
$\frac{d}{dx}(\frac{x^{3}+x^{2}+1}{x^{4}+2x+1})$

Here, the differentiation is with respect to x
And the variable x is in both numerator and denominator.

We have,
$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{g^{2}(x)}$ -----[1]

Here, $f'(x) = \frac{d}{dx}(f(x))$
and $g'(x) = \frac{d}{dx}(g(x))$

So,
we take
$f(x) = x^{3}+x^{2}+1$
$g(x) = x^{4}+2x+1$

$f'(x) = \frac{d}{dx}(x^{3}+x^{2}+1)$
By Applying linearity of differentiation
$\implies f'(x) = 3x^{2} + 2x + 0 = 3x^{2}+2x$

$g'(x) = \frac{d}{dx}(x^{4}+2x+1)$
By Applying linearity of differentiation
$\implies g'(x) = 4x^{3} + 2(1) + 0 = 4x^{3}+2$

Substitute in formula

$\implies \frac{(x^{4}+2x+1)(3x^{2}+2x)-(x^{3}+x^{2}+1)(3x^{2}+2x)}{(x^{4}+2x+1)^{2}}$

$\implies \frac{(3x^{6}+2x^{5}+6x^{3}+4x^{2}+3x^{2}+2x)-(3x^{5}+2x^{4}+3x^{4}+2x^{3}+3x^{2}+2x)}{(x^{4}+2x+1)^{2}}$

After simplifying,

$\frac{(3x^{6}-2x^{5}-5x^{4}+4x^{3}+4x^{2})}{(x^{4}+2x+1)^{2}}$