SO, I have been given: y^(2) = x^(2) / (xy – 28) at (8,4).

I am clueless on how to do this. In m calc group, I cannot find a problem as similar as this to use an example. Can someone show me step-by-step on how to get to the derivative in order to plug in (8,4)?

6205 says

Slope of the tangent is the value of (dy/dx) at (8, 4)

y^2 = x^2 / (xy – 28)

=> y^2 (xy – 28) = x^2

=> xy^3 – 28y^2 = x^2

=> y^3 + 3xy^2 dy/dx – 56y dy/dx = 2x

Plugging x = 8 and y = 4

=> 4^3 + 3 * 8 * 4^2 dy/dx – 56 * 4 * dy/dx = 2 * 8

=> 64 + 384 dy/dx – 224 dy/dx = 16

=> 160 dy/dx = – 48

=> dy/dx = – 3/10.

123 says

Slope of tangent is dy/dx

So you just differentiate the given equation, and put y=4 and x=8

Cross multiplying…

x(y)^3 – 28(y)^2 = x^2

Differentiating wrt x…. Applying product rule…[ d/dx(uv)= u dv/dx+v du/dx ]

3* (y)^2* x* dy/dx + (y)^3 – 28(2y) dy/dx= 2x

Grouping like terms…

dy/dx( 3(y)^2*x- 56y ) +64 = 2x

Substitute values of x and y…

dy/dx(3*16*8 – 56*4)= 16-64

dy/dx= -48/160

Slope = -3/10

2 Beautiful Daughters8928 says

To find the derivative we must use implicit rule… find the derivative of each term in relation to x…

Note: the derivative in relation to x of y^2 is 2y*y´ … chain rule. In right side we must use quotient rule.

Then it is 2y*y´ = [2x(xy-28) – x^2*(1y+xy´)]/(xy-28)^2 …. now put x= 8 and y = 4 and find y´

16y´ = [16(32-28) -64*(4 +8y´)] /(32 – 28) ==> we can divide each term by 16

y´= [ 4 – 16 -32y´]/4 ==> y´= 1- 4 – 8y´ ==> 9y´= -3 ==> y´ = -1/3 OK!

2292 says

Ok So… no idea if this is right, but its worth a shot…

Equation:

y^2 = x^2 / (xy-28)

Derivative:

d/dx(y(x)^2) = d/dx(x^2/(x y(x)-28))

Using the chain rule:

2 y(x) (d/dx(y(x))) = d/dx(x^2/(x y(x)-28))

The derivative of y(x) is y'(x):

2 y(x) y'(x) = d/dx(x^2/(x y(x)-28))

Use the quotient rule:

2 y(x) y'(x) = ((x y(x)-28) (d/dx(x^2))-x^2 (d/dx(x y(x)-28)))/(x y(x)-28)^2

Differentiate the sum term by term:

2 y(x) y'(x) = ((x y(x)-28) (d/dx(x^2))-x^2 (d/dx(x y(x))+d/dx(-28)))/(x y(x)-28)^2

The derivative of -28 is zero:

2 y(x) y'(x) = ((x y(x)-28) (d/dx(x^2))-x^2 (d/dx(x y(x))+0))/(x y(x)-28)^2

Use the product rule:

2 y(x) y'(x) = ((x y(x)-28) (d/dx(x^2))-x^2 (x (d/dx(y(x)))+y(x) (d/dx(x))))/(x y(x)-28)^2

The derivative of y(x) is y'(x):

2 y(x) y'(x) = ((x y(x)-28) (d/dx(x^2))-x^2 (y(x) (d/dx(x))+x y'(x)))/(x y(x)-28)^2

The derivative of x^2 is 2 x:

2 y(x) y'(x) = ((2 x) (x y(x)-28)-x^2 (y(x) (d/dx(x))+x y'(x)))/(x y(x)-28)^2

Final Derivative:

(2 x (x y(x)-28)-x^2 (x y'(x)+1 y(x)))/(x y(x)-28)^2

Pluging in x and y, slope of tangent = 2.75

Equation of Tangent:

y -yo = m(x -xo)

y-4 = 2.75(x-8)

y-4 = 2.75x-22

y = 2.75x -18