What is the slope of the tangent line of $(x y - \frac{1}{y x}) (x y + \frac{1}{x y}) = C$ $(xy-1/(yx))(xy+1/(xy)) =C$ , where C is an arbitrary constant, at $(1, 5)$ $(1,5)$ ?

Question

What is the slope of the tangent line of $(x y - \frac{1}{y x}) (x y + \frac{1}{x y}) = C$ $(xy-1/(yx))(xy+1/(xy)) =C$ , where C is an arbitrary constant, at $(1, 5)$ $(1,5)$ ?

1 Answer

Impact of this question

1440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-15T01:58:05

I got $- 5$ $-5$

Explanation:

We are told that the point $(1, 5)$ $(1,5)$ lies on $(x y - \frac{1}{y x}) (x y + \frac{1}{x y}) = C$ $(xy-1/(yx))(xy+1/(xy)) =C$ .

So we can find $C$ $C$ to be $25 - \frac{1}{25}$ $25-1/25$ . (We won't actually need this.)

As written, this looks tedious to differentiate implicitly, so let's rewrite it as:

$x^{2} y^{2} - \frac{1}{x^{2} y^{2}} = C$ $x^2y^2-1/(x^2y^2) = C$ .

Or better, still,

$x^{4} y^{4} - 1 = C x^{2} y^{2}$ $x^4y^4-1=Cx^2y^2$ . So

$4 x^{3} y^{4} + 4 x^{4} y^{3} \frac{d y}{d x} = 2 C x y^{2} + 2 C x^{2} y \frac{d y}{d x}$ $4x^3y^4+4x^4y^3 dy/dx = 2Cxy^2+2Cx^2y dy/dx$

$2 x^{2} y^{3} + 2 x^{3} y^{2} \frac{d y}{d x} = C y + C x \frac{d y}{d x}$ $2x^2y^3+2x^3y^2 dy/dx = Cy+Cx dy/dx$

Solving for the derivative gets us

$\frac{d y}{d x} = - \frac{2 x^{2} y^{3} - C y}{2 x^{3} y^{2} - C x}$ $dy/dx = -(2x^2y^3-Cy)/(2x^3y^2-Cx)$

Evaluating at $(1, 5)$ $(1,5)$ , we get

$\frac{d y}{d x} {⎤ ⎥ ⎦}_{(1,5)} = - \frac{2 {(5)}^{3} - 5 C}{2 {(5)}^{2} - C} = - \frac{5 (2 {(5)}^{2} - C)}{2 {(5)}^{2} - C} = - 5$ $dy/dx]_"(1,5)" = -(2(5)^3-5C)/(2(5)^2-C) = -(5(2(5)^2-C))/(2(5)^2-C)=-5$

Bonus

Regardless of the constant, at $x = 1$ $x=1$ , the slope of the tangent will be $- y$ $-y$

Science

Math

Humanities

... and beyond

What is the slope of the tangent line of $(x y - \frac{1}{y x}) (x y + \frac{1}{x y}) = C$ $(xy-1/(yx))(xy+1/(xy)) =C$ , where C is an arbitrary constant, at $(1, 5)$ $(1,5)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the slope of the tangent line of (xy-1/(yx))(xy+1/(xy)) =C (xy−1yx)(xy+1xy)=C (xy-1/(yx))(xy+1/(xy)) =C , where C is an arbitrary constant, at (1,5)(1,5)(1,5)?

1 Answer

Explanation:

Related questions

Impact of this question

What is the slope of the tangent line of $(x y - \frac{1}{y x}) (x y + \frac{1}{x y}) = C$ $(xy-1/(yx))(xy+1/(xy)) =C$ , where C is an arbitrary constant, at $(1, 5)$ $(1,5)$ ?