Please show your result for implicit differentiation prior to substituting for the given point. I get something different for y' at that point, but it's similarly complicated.kylem1994 said:Homework Statement
Find the equation of the tangent line of tan(xy2)=(2xy)/[itex]\pi[/itex] at (-[itex]\pi[/itex],1/2)
Homework Equations
The Attempt at a Solution
I managed to get the equation into its dy/dx form and for the slope to be (1-.5pi)/(2pi-2pi2)
This seems far to complicated to be correct though.. can someone confirm this?
I checked your (implicit) differentiation and the value of the derivative at (-π, 1/2) with WolframAlpha, and it agrees with your results totally.kylem1994 said:Here's what I got, dy/dx = [y2(pi)sec2(xy2)-2y]/[(2x)(1-y(pi)sec2(xy2)]
SammyS said:I checked your (implicit) differentiation and the value of the derivative at (-π, 1/2) with WolframAlpha, and it agrees with your results totally.
Yes, I'm quite sure that it's correct.kylem1994 said:That's always good to hear :p So my slope I found is most likely right? Even tho its a mess to look at?
Implicit differentiation is a technique used in calculus to find the derivative of an equation that is not explicitly written in terms of one variable. In this case, the equation is in terms of both x and y, so we use implicit differentiation to find the derivative with respect to x.
To find the derivative of tan(xy^2), we use the chain rule. First we take the derivative of the outer function, which is tan(u), where u=xy^2. This gives us sec^2(xy^2). Then we multiply by the derivative of the inner function, which is y^2. So the final derivative is 2xysec^2(xy^2).
The equation of the tangent line at a specific point can be found by plugging the x and y values of the point into the derivative we found in the previous step. This will give us the slope of the tangent line. Then, using the point-slope form of a line, we can find the equation of the tangent line.
Yes, we can find the equation of the tangent line at any point on the curve by following the steps outlined in the previous two questions. However, it is important to keep in mind that the equation of the tangent line will change depending on the point chosen.
We use implicit differentiation in this problem because the equation is not explicitly written in terms of one variable. This makes it impossible to use the standard techniques for finding the derivative, such as the power rule or product rule. Implicit differentiation allows us to find the derivative of equations that are more complex and involve multiple variables.