Question

differentiate f(x)=(x+10xsqrtx)/(sqrtx)

Answer 1

@SolomonZelman \(x^{-1/2}\)

Answer 2

???

Answer 3

look at the original post, it is over sqrt x

Answer 4

@deannapartida , follow the method proposed by @SolomonZelman, but you need to determine what \(\frac{1}{\sqrt x}\) is as an exponent

Answer 5

my bad

Answer 6

it's cool, but you can explain, you were here first :)

Answer 7

\[f \left( x \right)=\frac{ x+10 x \sqrt{x} }{ \sqrt{x} }=\frac{ x }{ \sqrt{x} }+\frac{10x \sqrt{x} }{ \sqrt{x} }= \sqrt{x}+10x\]
\[\frac{ d }{ dx }\left( \sqrt{x} \right)=\frac{ 1 }{ 2 \sqrt{x} }\]

Answer 8

\[f(x)=\frac{(x+10x\sqrt{x})}{(\sqrt{x})}\]\[f(x) = (x+10x^{1+1/2}) \cdot x^{-1/2}\]\[f(x) = \color{red}{(x+10x^{3/2})}\cdot \color{blue}{x^{-1/2}}\]\[f'(x) = \color{red}{f'}\color{blue}{g} +\color{blue}{g'}\color{red}f\]

Answer 9

I combined \(10x\cdot \sqrt{x}\) as \(10x \cdot x^{1/2} \implies 10x^{1+1/2} = 10x^{3/2}\)

Then instead of using the quotient rule to differentiate the function, I'm using the product rule because I think it's easier, (for me atleast)

Answer 10

You actually don't need the product rule if you distribute :)