Using implicit differentiation, find the equation of the tangent line to the given curve at the given point: 3x2y2 − 3y −17 = 5x +14 at (1,−3)

Here's a simple example comparison.

Non-implicit ( y = f(x) )

y = x + 1          

dy/dx = 1

Implicit ( y = f(x,y) )

y = x*y + 1

dy/dx = y*1 + x*dy/dx

(1 - x)dy/dx = y

dy/dx = y/(1 - x)

But (1 - x)y = 1, or y = 1/(1 - x), so, substituting this in the above, we get dy/dx = 1/(1-x)²

In this case, from y = 1/(1-x) we could have written directly;  dy/dx = -(1-x)^-2*(-1)  or  dy/dx = 1/(1-x)², the same as above.  

However, we can't always rearrange y = f(x,y) to get y = f(x) explicitly, so sometimes we are forced to use implicit differentiation.

.

3x^2y^2 − 3y −17 = 5x +14

6xy^2 + 6x^2yy ' - 3y '  = 5

y' ( 6x^2y - 3)  =  5 - 6xy^2

y' =   ( 5 - 6xy^2) / (6x^2y - 3)        and the slope at (1, -3)  = [(5 - 6(1)(-3)^2] / [(6(1)^2(-3) - 3) ] = -49 /  -21   =   7/3

And the equation of the tangent line at this point is.....

y = (7/3)(x - 1) -3  

y =(7/3)x - (7/3) - 3

y = (7/3)x - 16/3

Here's a graph...........https://www.desmos.com/calculator/qf9zrfot56

Using implicit differentiation, find the equation of the tangent line to the given curve at the given point: 3x2y2 − 3y −17 = 5x +14 at (1,−3)

I. implicit differentiation:

$$\small{\text{$
\begin{array}{rcl}
3x^2y^2 - 3y -17 &=& 5x +14 \\
f(x,y) &=& 3x^2y^2 - 3y-5x -31 = 0\\\\
f'(x,y) &=& -\dfrac{ \dfrac{ d( 3x^2y^2 - 3y-5x -31 ) } {dx} }
{ \dfrac{ d( 3x^2y^2 - 3y-5x -31 ) } {dy} }\\\\\\
f'(x,y) &=& -\dfrac{ 6xy^2-5 }
{ 3x^2\cdot 2y-3 }\\\\\\
f'(x,y) &=& \dfrac{ 5- 6xy^2 }
{ 6yx^2-3 }
\end{array}
$}}$$

II.  tangent line

$$\small{\text{$
\begin{array}{rcl}
x_p=1 \qquad y_p = -3\\\\
f'(x_p,y_p) &=& \dfrac{y-y_p}{x-x_p}\\\\
y &=& f'(x_p,y_p)\cdot(x-x_p)+y_p \qquad | \qquad f'(x_p,y_p) = \dfrac{5-6\cdot 1\cdot (-3)^2} {6\cdot (-3)\cdot 1^2 -3 } = 2.\overline{3}\\\\
\mathbf{y} &\mathbf{=}&\mathbf{ 2.\overline{3} \cdot(x-1)-3}\\\\
\mathbf{y} &\mathbf{=}&\mathbf{\dfrac{7}{3} \cdot(x-1)-3}
\end{array}
$}}$$

Thanks Chris and Heureka  

What sort of differentiation is implicit?  What does implicit mean in this context?

i'm sure heureka (or Alan or Bertie) can explain this better, but "implicit"  means that we can't "directly" take the derivative of this function because there is no way to separate the x and y terms so that we can isolate y - as we might normally do.

So....we use the Chain/Product Rules  by treating y as an "unknown" function of x....... and we can differentiate with respect to "x," first, and then with respect to "y".....  

This allows us to "collect" all the terms that have y '  - (or.....dy/dx) - associated with them and then "solve" for y'.........

(I tend to use y ' for dy/dx.......whatever.......this procedure has always seemed to me to have a bit of " black magic" about it.......LOL!!!!  }

Oh ok thanks Chris :)

Thanks Alan   

?

It comes from the word implied.

There is an implied relationship  between the two variables.

It may or may not be possible (or desirable) to convert it into an explicit relationship.

Thanks Bertie