derivative of y = (sqrt x) - (1/2x) + (x/sqrt 2)

[Becky4paws]

derivatives with exponents, fractions and square roots (AAAGGGHH).

y = (sqrt x) - (1/2x) + (x/sqrt 2)

I hope that notation is clear enough for you...the last part is x/square root of 2.

y = (x^1/2) - (1/2x^-1) + (x/2^-1/2)
y' = (1/2x^-1/2) - (-1)(-1/2)(x^-2) + x(-1/2)(2^-3/2)
y' = (1/2x^-1/2) - (1/2x^-2) + x (-1^-3/2)
y' = [1/(2sqrt x)] - [1/(2x^2)] + [x/(cubert -1)]

This one isn't easy... how did I do?

 

[galactus]

\(\displaystyle \L\\\sqrt{x}-\frac{1}{2x}+\frac{1}{\sqrt{2}}x\)

Take it one at a time:

\(\displaystyle \L\\\sqrt{x}=x^{\frac{1}{2}}\)

\(\displaystyle \L\\\frac{d}{dx}[x^{\frac{1}{2}}]=\frac{1}{2}x^{\frac{-1}{2}}=\frac{1}{2\sqrt{x}}\)


\(\displaystyle \L\\\frac{d}{dx}[\frac{1}{2x}]=\frac{d}{dx}[\frac{1}{2}x^{-1}]=\frac{-1}{2}x^{-2}=\frac{-1}{2x^{2}}\)


\(\displaystyle \L\\\frac{d}{dx}[\frac{1}{\sqrt{2}}x]=\frac{1}{\sqrt{2}}\)


Put them together.