Take the derivative of both sides of the equation implicitly:
At x=7, y=1. Thus, 1^3+3(7)(1^2)*y'+1+7y'=0
which yields y'=-2/28=-1/14
The slope of the tangent line is m which also is y'. Thus, you now have m=-1/14. Since the point (7,1) is on the tangent line, then just plug these coordinates as well as the value of m into the equation y=mx+b and solve for b.
Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=5 at the point (4,1) . The equation of this tangent line can be written in the form y = mx+b where m is:? and where b is:?
I am unsure of how to take the derivative of this equation. It may be the exponents that are giving me trouble but I'm not sure exactly. Find the equation of the tangent line to the curve 4e^xy = 2x + y at point (0,4). On the left