Differentiate both sides with respect to x

$\frac{d}{dx}(x^{\frac{1}{2}} + y^{\frac{1}{2}}) = \frac{d}{dx}(1)\\\\ \frac{1}{2x^{\frac{1}{2}}} + \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0\\\\\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}\\\\\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \\\\\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \times 2\sqrt{y}\\\\\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \\\\\boxed{\bf{y'= -\frac{\sqrt{y}}{\sqrt{x}}}}$

5 0

2 years ago

The product of 8/15, 6/5,1/3 is

Debora [2.8K]

Product is the result of multiplying the numbers, so the product of 8/15, 6/5,1/3 means that we have to multiply the three numbers, like this:

$\frac{8}{15} * \frac{6}{5} * \frac{1}{3}$
let's make them into one fraction:

$\frac{8*6*1}{15*5*3}$

We can simplyfy:

 $\frac{8*6}{15*5*3}$


 $\frac{8*2}{15*5}$

and multiply:

 $\frac{16}{75}$

and this is the correct answer!

3 0

3 years ago