Question

Differentiate w r yo 'x'
y=(x^(3)+1)/(x+1)

Answer 1

$\huge \boxed{ \red{ \mathfrak{solution}}}$

Given:

$\bold{y = \frac{ {x}^{3} + 1}{x + 1} }$

First solve y.

$\star \: y = \frac{ {x}^{3} + 1}{x + 1} \\ \\ \rightarrow \: y = \frac{ (x + 1)(x {}^{2} - x + {1}^{2} )}{x + 1} \\ \\ \star \: y = \frac{ \cancel{(x + 1)}( {x}^{2} - x + 1)}{ \cancel{x + 1}} \\ \\ \rightarrow \: y = {x}^{2} - x + 1$

Now differentiate w.r.t X both sides,

$\star \: \frac{dy}{dx} = \frac{d}{dx} ( {x}^{2} - x + 1) \\ \\ \star \: \frac{dy}{dx} = 2 {x}^{2 - 1} - 1 {x}^{1 - 1} + 0 \\ \\ \star \frac{dy}{dx} = 2x - 1$

Formula used :

$\red \star \boxed{ \bold{ {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )}}$

$\green \star \: \bold{\frac{d}{dx} ( {x}^{y} ) = y {x}^{y - 1} }$

$\pink\star \boxed{\bold{\frac{d}{dx} (constant) = 0}}$