Need help urgently

Question

1. Find the equation of the line tangent to the curve 𝑦 = 1 √𝑥2+2 at 𝑥 = −1. Do not use the derivatives rules. Use the limit definition to calculate the derivative. [5 marks]

need solution urgently

Answer 1

slope = y' = the derivative of y

not sure what y is though? that 1 is confusing. As is how far the square root sign extends. Maybe it's supposed to be y= 1/sqr(x^2 + 2). = (x^2+2)^-1/2

Then y' = (-1/2)(x^2+2)^-3/2

plug in x=-1

y' = (-1/2)(1+2)^-3/2 = -1/(2)(3)^3/2

= -1/2(sqr27) =-1/6sqr3

At x=-1 y = 1/sqr1 = 1/1 = 1

the tangent line has slope -1/6sqr3 and goes through (-1,1)

y=(-1/6sqr3)x + b, plug in x=-1 and y=1

1 =(-1/6sqr3)(-1) + b = 1/6sqr3 + b

b = 1-1/6sqr3

tangent line is

y=-(1/6sqr3)x + 1 - 1/6sqr3 or

6sqr3 y= -x + 1/6sqr3 - 1 or with sqr3 = about 1.7

10.2y = -x - .9