Equation of the tangent to the curve `y=e^(x) at (0,1)` is

Equation of the tangent to the curve y= log x at (1,0) is

The equation of the tangent to the curve y=4xe^(x) at (-1,(-4)/e) is

The equation of the tangent to the curve y=e^(-|x|) at the point where the curve cuts the line x = 1, is

The equation of the tangent to the curve y=4e^(-x/4) at the point where the curve crosses Y-axis is equal to

Equation of the tangent to the curve y= e^(-abs(x)) at the point where it cuts the line x=1-

Find the equation of the tangent of the curve y=3x^(2) at (1, 1).

The equation of the tangent to the curve y=1-e^(x//2) at the tangent to the curve y=1-e^(x//2) at the point of intersection with the y-axis, is

Equation of the tangent to the curve y=1-e^((x)/(2)) at the point where the curve cuts y -axis is

The equation of the tangent to the curve y=1-e^((x)/(2)) at the point of intersection with the y -axis,is