How do you find the Tangent line to a curve by implicit differentiation?

Question

Elijah Abram · Accepted Answer

Let us this example: Find the equation of the tangent line to the circle #x^2+y^2=5^2# at the point #(3,4)#.  In order to identify a line, we need two pieces of information:
# {("Point: " (x_1,y_1)=(3,4)), ("Slope: " m=?):}# Since the point is already provided, all you need is the slope #m#.
Let us find #m# by implicit differentiation. By implicitly differentiating, #d/{dx}(x^2+y^2)=d/{dx}(5^2)Rightarrow 2x+2y{dy}/{dx}=0# by dividing by #2y#, #{x}/{y}+{dy}/{dx}=0# by subtracting #x/y#, #{dy}/{dx}=-x/y# So, we can find #m# by evaluating #{dy}/{dx}# at #(3,4)#. #m={dy}/{dx}|_{(3,4)}=-3/4# By Point-Slope Form: #y-y_1=m(x-x_1)#, Tangent Line: #y-4=-3/4(x-3)#

Evelyn Agard · Answer

undefined To find the tangent line to a curve by implicit differentiation, follow these steps:

1. Differentiate both sides of the equation with respect to the variable in question.
2. Treat the derivative of the dependent variable as a function of the independent variable.
3. Solve the resulting equation for the derivative of the dependent variable.
4. Substitute the values of the independent and dependent variables into the derivative expression to find the slope of the tangent line.
5. Use the point-slope form of a line to write the equation of the tangent line, using the slope found in the previous step and the coordinates of the point of tangency.