Question

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

Answer 1

chain rule for this one

Answer 2

\[\frac{d}{dx}(x+\sqrt{x})^{-2}=-2(x+\sqrt{x})^{-3}\times (1+\frac{1}{2\sqrt{x}})\]

Answer 3

I don't know how to get there

Answer 4

ok it is the chain rule, the power rule, and the fact that i have memorized
\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\]

Answer 5

I know that you bring -2 out front, but where does the ^-3 come from? or the second half? I've missed a week of school.

Answer 6

you have a composite function. i looks like
\[(f(x))^{-2}\] were
\[f(x)=x+\sqrt{x}\] the derivative of something to the power of minus 2 is minus 2 something to the power of -3 (by the power rule) times the derivative of "something"

Answer 7

\[\frac{d}{dx}f(x)^n=nf^{n-1}(x)\times f'(x)\]

Answer 8

just like
\[\frac{d}{dx}x^n=nx^{n-1}\]

Answer 9

so to answer more specifically, the -3 comes from -2 - 1

Answer 10

ohh so the second half is just the derivative?

Answer 11

the second half is the derivative of the "inside function" yes. and the derivative of
\[x+\sqrt{x}\] is
\[1+\frac{1}{2\sqrt{x}}\]

Answer 12

okay so every time i have a \[\sqrt{x}\] , then it's \[1/2\sqrt{x}\]?