y=3x^(4)-4x" at "x=1

The two curves y=x^2+1,y=3x^2-4x+3 at (1,2)

Find the equation of the tangent line to the curve y=(3x^2-1)/x at the point (1,2) is (a) 4x-y-1=0 (b) 4x+y+2=0 (c)x-y-1=0 (d) 4x-y-2=0

If y = cos^(-1) (4x^(3) -3x) " then " (dy)/(dx)= ?

If (1)/(x)+(1)/(y)=(1)/(4) and (1)/(x)-(1)/(y)=(3)/(4) , then x=

Find second order derivative for y=3x^4+4x^2-x+1

If x^(3)=y^(4) prove that (x/y)^(4/3)+(y/x)^(3/4)=x^(1/3)+y^(-1/4)

When 0 lt x lt 1 , y=1/2x^2+2/3x^3+3/4x^4+ . . . Then the value of e^(1-y) at x=1/2

Find (x+y)+(x-y), if x=(5)/(4),y=(-1)/(3) (ii) x=(2)/(7),y=(4)/(3)( iii) x=(1)/(4),y=(3)/(2)

The value of f(x,y)=((x^(3)y)^((1)/(4))-(y^(3)x)^((1)/(4)))/(sqrt(y)-sqrt(x))+(1+sqrt(xy))/((xy)^((1)/(4)))xx((1+2sqrt((y)/(x))+(y)/(x))^(4) when x=9 and y=0.04