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# 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)

+5
2045
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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)

Jun 11, 2015

#6
+32236
+23

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.

.

Jun 11, 2015

#1
+119758
+15

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

Jun 11, 2015
#2
+25911
+18

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

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

Jun 11, 2015
#3
+113567
0

Thanks Chris and Heureka

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

Jun 11, 2015
#4
+119758
+18

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!!!!  }

Jun 11, 2015
#5
+113567
+5

Oh ok thanks Chris :)

Jun 11, 2015
#6
+32236
+23

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.

.

Alan Jun 11, 2015
#7
+113567
+5

Thanks Alan

Jun 11, 2015
#8
+890
+10

?

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.

Jun 11, 2015
#9
+113567
+5

Thanks Bertie

Jun 12, 2015