Question

Find dx/dy of y=(x+sqrt(x))^-2

Answer 1

Are you finding dx/dy or dy/dx?

Answer 2

dy/dx sorry, miss type

Answer 3

Did you apply chain rule?

Answer 4

I don't understand how to use the chain rule

Answer 5

what if you were to just differentiate x^(-2) w.r.t x?

Answer 6

could you do that

Answer 7

-2x^-3

Answer 8

\[\frac{dy}{dx}=-2(x+\sqrt{x})^{-3} \cdot (x+\sqrt{x})'\]
So now you just need to find the derivative of the inside

Answer 9

\[\frac{d}{dx}(x+\sqrt{x})=?\]

Answer 10

\[\frac{d}{dx}(x+\sqrt{x})=\frac{d}{dx}(x+x^\frac{1}{2})=?\]

Answer 11

would it be...\[1+\frac{ 1 }{ 2\sqrt{x} }\]

Answer 12

yes

Answer 13

\[\frac{dy}{dx}=-2(x+\sqrt{x})^{-3} \cdot (x+\sqrt{x})' \\ \frac{dy}{dx}=-2(x+\sqrt{x})^{-3} (1+\frac{1}{2 \sqrt{x}}) \]

Answer 14

ok thank you!