Find the slope of the tangent line to the curve defined by 9x^3+4xy−6y^2=1752 at the point (6,−4).

Question

This is a derivative but I don't know how to do this question! I would appreciate if all work is shown with detailed steps!

Provide the answer to the slope of the curve at the point (6,-4) is_____.

Answer 1

The slope of the tangent line is the value of the derivative of the function at the point in question. Thus ...

f = 9x^3+4xy-6y^2=1752 ... f'=27x^2+4xdy/dx+4y-12ydy/dx=0 ... solve for dy/dx = (-27x^2-4y)/(4-12y) ... next evaluate dy/dx at (6,-4), dy/dx= (-27*36-+16)/(4+16)= 47.8