AIATS DOUBTS

Q4. Make it an implicit function.

yx² + yx + y = x³ + 2x² + 2x + 1

We can make use of partial differentiation now.
Differentiate the function wrt x treating y as constant.

f_x = 2xy +y + 0 - 3x² - 4x -2 - 0
Similarly, differentiate the function wrt y treating x as constant.

f_y = x² + x + 1 - 0 - 0 - 0 - 0

Now by partial differentiation formula,
dydx = -f_xf_y
= 3x²+ 4x -2xy - y + 2x² + x + 1

Very useful for implicit functions. I don't know if it could be used for nth derivative however...as mostly we get questions asking for d²ydx².

So at x = 1,
dydx = 3 + 4 - 2(2) - 2 + 23 = 1.
:)

Q3. f(x) = x.cosx
f'(x) = cosx - xsinx

Is the limit undefined? I must be making a mistake there.

Q1. (x-1)(x-6)/(x-2)(x-4) = e^x

Lim_{x→∞}(x-1)(x-6)/(x-2)(x-4) = 1 (Division of highest degree co-efficients)
So when x tends to infinity, curve's output is 1.
Also, we find the sign scheme of the polynomial by wavy curve method.

I think it has only one solution..