# Thread: A question related to Derivative Rules.

1. ## A question related to Derivative Rules.

Hey everyone.

I'm currently trying to solve a problem from my textbook that was assigned for homework. It reads;

"Let P(a, b) be a point on the curve sqroot(x) + sqroot(y) = 1. Show that the slope of a tangent at P is -sqroot(b/a)."

Using Graphmatica, (1/x^2)+(1/y^2)=1 shows four asymptotes, one in each quadrant, all approaching 1 or (-1) (depending on the positive/negative side of the y axis).

How abouts do I solve this?

2. Originally Posted by Jeavus
Hey everyone.

I'm currently trying to solve a problem from my textbook that was assigned for homework. It reads;

"Let P(a, b) be a point on the curve sqroot(x) + sqroot(y) = 1. Show that the slope of a tangent at P is -sqroot(b/a)."

Using Graphmatica, (1/x^2)+(1/y^2)=1 shows four asymptotes, one in each quadrant, all approaching 1 or (-1) (depending on the positive/negative side of the y axis).

How abouts do I solve this?

Are you asking about two different problems? I don't see how $\displaystyle \sqrt{x}+\sqrt{y}=1$ relates to $\displaystyle \frac{1}{x^2}+\frac{1}{y^2}=1$.

3. Oh, I'm sorry. I meant to write x^1/2, not 1/x^2.

4. Originally Posted by Jeavus
Hey everyone.

I'm currently trying to solve a problem from my textbook that was assigned for homework. It reads;

"Let P(a, b) be a point on the curve sqroot(x) + sqroot(y) = 1. Show that the slope of a tangent at P is -sqroot(b/a)."

Using Graphmatica, (1/x^2)+(1/y^2)=1 shows four asymptotes, one in each quadrant, all approaching 1 or (-1) (depending on the positive/negative side of the y axis).

How abouts do I solve this?

$\displaystyle x^{\frac{1}{2}}+y^{\frac{1}{2}}=1$

The derivative would be (using implicit differentiation):

$\displaystyle \frac{1}{2}x^{-\frac{1}{2}}dx+\frac{1}{2}y^{-\frac{1}{2}}dy=0$

$\displaystyle \frac{1}{2\sqrt{x}}dx+\frac{1}{2\sqrt{y}}dy=0$

$\displaystyle \frac{1}{2\sqrt{y}}dy=-\frac{1}{2\sqrt{x}}dx$

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{x}{y}}$

Plugging in the point P(a,b), you'd get:

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{a}{b}}$

5. Thanks very much.

I can see where I was thinking wrong.

6. Originally Posted by ecMathGeek
$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{x}{y}}$

Plugging in the point P(a,b), you'd get:

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{a}{b}}$

I made a typo. This should have been:

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$

Which, for P(a,b), would give us:

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{b}{a}}$