Question

find dy/dx : x√1+x³ at X =2​

Answer 1

Step-by-step explanation:

We have,

$y = x \sqrt{1 + {x}^{3} }$

$= > \frac{dy}{dx} = \sqrt{1 + {x}^{3} }. \frac{d}{dx} (x) + x. \frac{d}{dx} ( \sqrt{1 + {x}^{3} })$

$= > \frac{dy}{dx} = \sqrt{1 + {x}^{3} } + x. \frac{3 {x}^{2} }{2 \sqrt{1 + {x}^{3} } }$

$= > ( \frac{dy}{dx}) _{x = 2} = \sqrt{1 + {2}^{3} } + \frac{3(2)^{3} }{2 \sqrt{1 + {2}^{3} } }$

$= > ( \frac{dy}{dx}) _{x = 2} = \sqrt{1 + 8 } + \frac{3 \times 8 }{2 \sqrt{9 } }$

$= > ( \frac{dy}{dx}) _{x = 2} = 3 + \frac{3 \times 8}{2 \times 3}$

$= > ( \frac{dy}{dx}) _{x = 2} = 7$