Double check my work (find rate of change)

[Becky4paws]

this one is tough for me.

Find the rate of change for the given function f(x) with respect to x for the prescribed value of x.

f(x) = (x + square root x)/(x*square root x)

f'(x) = (1 + x^1/2)/(1*x^-1/2) = 1 + 1/2*x^-1/2/(1*-1/2*x^-3/2)

 

[guntram]

what does it mean "rate of change"

f'(x)=2/(x^1/2)-3/2

 

[soroban]

Re: Double check my work

Hello, Becky4paws!

Find the rate of change for the given function \(\displaystyle f(x)\) with respect to \(\displaystyle x\) for the prescribed value of \(\displaystyle x.\)

\(\displaystyle \L\;\;\;f(x) \:= \:\frac{x\,+\,\sqrt{x}}{x\cdot\sqrt{x}}\)

Why not simplify first?

We have: \(\displaystyle \L\,f(x)\;=\;\frac{x\,+\,x^{\frac{1}{2}}}{x\cdot x^{\frac{1}{2}}} \;=\;\frac{x\,+\,x^{\frac{1}{2}}}{x^{\frac{3}{2}}} \;= \;\frac{x}{x^{\frac{3}{2}}}\,+\,\frac{x^{\frac{1}{2}}}{x^{\frac{3}{2}}}\)

Hence: \(\displaystyle \L\,f(x)\;=\;x^{-\frac{1}{2}}\,+\,x^{-1}\)


Now differentiate . . .

 

[Becky4paws]

Another try

f(x)=x^-1/2 + x^-1 =
f"(x) = 1/square root x + 1/x