Question

the slope of the normal to the curve y= x^2- log x at x=2 is​

Answer 1

Step-by-step explanation:

We have,

$y = {x}^{2} - log(x)$

$= > \frac{dy}{dx} = 2x - \frac{1}{x}$

$= >( \frac{dy}{dx})_{x = 2} = 4 - \frac{1}{2} = \frac{7}{2}$

At x = 2, y = (2)² - log(2) = 4 - log(2)

Equation of normal :

$(y - (4 - log(2) )) \times \frac{7}{2} + (x - 2) = 0$

$= >7 (y - 4 + log(2) ) + 2(x - 2) = 0$

$= > 7y - 24 + 7 log(2) + 2x - 4 = 0$

$= > 7y + 2x + 7 log(2) - 28 = 0$