Question

y=((2x^1/2)/(2x^1/2+1))^1/2

Find dy/dx plz help me

Answer 1

What you need to find is the derivative. This is going to be a long and complicated problem. First we can use the power rule and change the problem to

1/2((2x^1/2)/(2x^1/2+1))^-1/2

Does that first step make sense? The power of 1/2 changed to -1/2 to compensate for 1/2 being brought to the front.

Answer 2

\[y ^{2}=\frac{ 2\sqrt{x} }{ 2\sqrt{x}+1 }\]
\[2y \frac{ dy }{ dx }=\frac{ \left( 2\sqrt{x}+1 \right)\frac{ 2 }{ 2\sqrt{x} }-2\sqrt{x}\frac{ 2 }{ 2\sqrt{x} } }{ \left( 2\sqrt{x}+1 \right)^{2}}\]
\[ \frac{ dy }{ dx }=\frac{ 1 }{ 2y }*\frac{ 2\sqrt{x}+1-2\sqrt{x} }{ \left( 2\sqrt{x}+1 \right)^{2} }\]
\[\frac{ dy }{ dx}=\frac{ 1 }{ 2y \left( 2\sqrt{x}+1 \right)^{2} }\]
substitute the value of y and simplify.