If `x^2+x y+y^2=7/4,` then `(dy)/(dx)` at `x=1` and `y=1/2` is:

If y^(2) -2x^(2) = y , then dy/dx at (1,-1) is

2 x y+y^2-2 x^2 (dy)/(dx)=0 , y=2 when x=1

(dy) / (dx) = (4x + y + 1) ^ (2)

If y= (sin^(-1) x)^2 , then (1-x^2) (d^2 y)/(dx^2) - x (dy/dx) = ..........

A: If y = x ^(y) then (dy)/(dx) = (y ^(2))/(x(1- log y )) If y = f (x) ^(y), then (dy)/(dx) = (y ^(2) f '(x))/(f (x) [1- ylog f (x)])= (y ^(2) f'(x))/(f (x) [1- log y])

If (sin x)^2 =x+y find (dy)/(dx) Find (dy)/(dx) if y=sin^(-1)[2^(x+1)/(1+4^x)]

If y=(1)/(x^(4)+x^(2)+1) , then find dy/dx

If y=(x^(7/2)+x^(-1/2))/(x^(7/2)-x^(-1/2)) , then (dy)/(dx)=

If y= (x^2+7) (x+1) , find dy/dx

If y= (x^2+7) (x+1) , find dy/dx