Find the slope of the tangent line to the curve at the point (3,2) . Give an exact value

Question

I asked this in one of my previous questions but I was wondering how to find the derivative of the tangent at a point.

the point is (3,2) and the problem states

Find the slope of the tangent line to the curve at the point. Give an exact value

I started by taking the derivative of the given function.

The function is x^3 + 5(x^2)y + 2y^2 = 4y + 117 

and the derivative I found is dy/dx= (-3x^2- 10xy) / (5x^2-4y-4)

This is where I am a little confused. In a previous question, what I understood, is that I find the derivative and evaluate the point by plugging it in?... I plugged in the x and y and I got -87/33 but the answer is wrong. If you could please explain where I misunderstood I would be so grateful.

Answer 1

Plug in x=3 and y=2 into the derivative you just found.  Then evaluate that derivative.  This will give you the slope of the tangent line.

 

Once you have the slope, use the point-slope form of the line to get your line:

 

y = (dy/dx)(x - x₁) + y₁

 

where   (x₁ , y₁) = (3, 2)

 

 

But if you got the wrong answer in your derivative, then you should double check your implicit differentiation.  See if you made an error.  Perhaps you did not apply the chain rule in the process.

 

 

As far as understanding, you quite understood the process.  Just check for any algebraic and differential errors.