Looks like there is a typo here.b521 said:Homework Statement
Find the equation of a tangent line at the curve at point (-3√3, 1)
x^(1/3) = y^(1/3) = 4
This would be the line with slope m, that passes through (1, 1).b521 said:Homework Equations
Point-slope:
y-1=m(x-1)
It's hard to tell whether this is correct, since the original equation likely has a typo. If you typed '=' in place of '+', then you have a sign error in your derivative. Also, it would be helpful to have an equation for your derivative; e.g., y' = y^(2/3)/x^(2/3).b521 said:The Attempt at a Solution
I took the derivative of that equation and resulted in
y^(2/3)/x^(2/3)
b521 said:When I tried plugging in x and y to get slope, the equation got very messy and I couldn't get a number out of it. Can anyone help me out?
Mark44 said:What is the original equation? I can't tell if your work is correct without knowing the original equation.
Mark44 said:(-3√3)^(2/3) can also be written as ((-3√3)^2)^(1/3)
Implicit differentiation is a method for finding the derivative of an equation that is not explicitly expressed in terms of one variable. This is useful for finding the slope of a tangent line to a curve at a specific point.
Regular differentiation involves finding the derivative of a function that is explicitly expressed in terms of one variable. Implicit differentiation, on the other hand, involves finding the derivative of a function that is not explicitly expressed in terms of one variable.
Implicit differentiation allows us to find the slope of the tangent line at a specific point on a curve without having to solve for one variable in terms of the other. This is particularly useful when the equation is difficult to solve for one variable or when the curve is not a function.
The first step is to differentiate both sides of the equation with respect to the independent variable. Then, solve for the derivative of the dependent variable. Finally, plug in the x-value of the point of interest to find the slope of the tangent line.
Implicit differentiation can only find the slope of the tangent line at a specific point on the curve. It cannot provide information about the curve as a whole or about other points on the curve. Additionally, it may be more complicated to use than regular differentiation for certain equations.