find the slope of the tangent line to y^4 = 3x^2 - 2x at the point (1,1)
4y^3 y' = 6x-2
y' = (6x-2)/(4y^3)
at (1,1) y' = 4/4 = 1
line is thus (y-1) = 1(x-1)
y = x
thank you for taking your time to answer my questions steve!
To find the slope of the tangent line to the curve at a given point, we can differentiate the equation of the curve with respect to x and then substitute the x-coordinate of the point into the resulting derivative expression.
Let's differentiate y^4 = 3x^2 - 2x with respect to x:
d/dx (y^4) = d/dx (3x^2 - 2x)
To differentiate y with respect to x, we can use implicit differentiation. Let u = y^4.
du/dx = d/dx (3x^2 - 2x)
Next, we can differentiate u = y^4 with respect to x:
du/dx = 4y^3 * dy/dx
Now, we can differentiate 3x^2 - 2x with respect to x:
du/dx = 6x - 2
Now we can substitute the x-coordinate of the given point (1,1) into the expression for du/dx:
du/dx = 6(1) - 2 = 6 - 2 = 4
So, the value of du/dx at (1,1) is 4. This represents the slope of the tangent line to the curve at that point. Therefore, the slope of the tangent line to y^4 = 3x^2 - 2x at the point (1,1) is 4.