Quotient Rule

Jan 2010
5
0
differentiate
(x^2+4x+3)/ X^(1/2)

here is what i have
x^(1/2) (2x+4) - (x^2 +4x +3) (1/2)x^-(1/2) all over x

and how do you write .5x^-.5 in radical form
 
Dec 2009
3,120
1,342
differentiate
(x^2+4x+3)/ X^(1/2)

here is what i have
x^(1/2) (2x+4) - (x^2 +4x +3) (1/2)x^-(1/2) all over x

and how do you write .5x^-.5 in radical form
Hi markwjak,

\(\displaystyle \frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2x^{\frac{1}{2}}}=\frac{1}{2\sqrt{x}}\)
 
Jan 2010
5
0
Hi markwjak,

\(\displaystyle \frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2x^{\frac{1}{2}}}=\frac{1}{2\sqrt{x}}\)
how would you simplify this

x^(1/2) (2x+4)- (x^2+4x+3) (1/(2x^(1/2))) over x

by dividing x^(1/2) and (1/(2x^(1/2))) from the equation?
 
Mar 2010
715
381
\(\displaystyle x^(1/2) (2x+4) - (x^2 +4x +3) (1/2)x^-(1/2) all over x \)
Correct (you might want to simplify that, though).
and how do you write .5x^-.5 in radical form
\(\displaystyle \dfrac{1}{2}\left(x\right)^{-\frac{1}{2}} = \dfrac{1}{2x^{\frac{1}{2}}} = \dfrac{1}{2\sqrt{x}}. \)
 
Last edited by a moderator:
Jan 2010
5
0
Wrong forum (it belongs to the calculus section, I think).
Correct (you might want to simplify that, though).

\(\displaystyle \dfrac{1}{2}\left(x\right)^{-\frac{1}{2}} = \dfrac{1}{2x^{\frac{1}{2}}} = \dfrac{1}{2\sqrt{x}}. \)
Thats the part I'm having trouble with is the simplifying
 
Dec 2009
3,120
1,342
how would you simplify this

x^(1/2) (2x+4)- (x^2+4x+3) (1/(2x^(1/2))) over x

by dividing x^(1/2) and (1/(2x^(1/2))) from the equation?
\(\displaystyle \frac{\sqrt{x}(2x+4)-\frac{\left(x^2+4x+3\right)}{2\sqrt{x}}}{x}\)

\(\displaystyle =\frac{\frac{2\sqrt{x}\sqrt{x}(2x+4)-\left(x^2+4x+3\right)}{2\sqrt{x}}}{x}\)

\(\displaystyle =\frac{2\sqrt{x}\sqrt{x}(2x+4)-\left(x^2+4x+3\right)}{2x\sqrt{x}}\)

\(\displaystyle =\frac{2x(2x+4)-\left(x^2+4x+3\right)}{2x\sqrt{x}}\)

\(\displaystyle =\frac{4x^2+8x-x^2-4x-3}{2x\sqrt{x}}=\frac{3x^2+4x-3}{2x\sqrt{x}}\)

that's about simplest
 
Mar 2010
715
381
Thats the part I'm having trouble with is the simplifying

\(\displaystyle \dfrac{\sqrt{x}\left(2x+4\right)-\dfrac{1}{2\sqrt{x}}\left(x^2+4x+3\right)}{x} = \dfrac{2\sqrt{x}\sqrt{x}\left(2x+4\right)-(x^2+4x+3)}{2\sqrt{x}x}\) \(\displaystyle = \dfrac{\left(2x(2x+4)-x^2-4x-3\right)}{2\sqrt{x^3}} = \dfrac{4x^2+8x-x^2-4x-3
}{2\sqrt{x^3}} = \boxed{\dfrac{3x^2+4x-3}{2\sqrt{x^3}}}.\)