Question

1Find dy/dx if (x √(a-x))/√(a +x )

Answer 1

Solution:-

$= \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]-[ \sqrt{a}-x]} * \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]+[ \sqrt{a}-x]}$

$= \frac{ [ \sqrt{a}+x]+[ \sqrt{a}-x]^{2} }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }$

$= \frac{ [ \sqrt{a}+x]^{2}+[ \sqrt{a}-x]^{2}+2[ \sqrt{a}+x][ \sqrt{a}-x] }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }$

$= \frac{4a}{4x \sqrt{a}}$

$= \frac{ \sqrt{a} }{x}$

Now if

$x=\frac{2ab}{1+ b^{2} }$

then we get

$\frac{ \sqrt{a} }{2ab}*[1+ b^{2}]$

$\frac{ [1+ b^{2}] }{2b \sqrt{a} }$

Answer 2

Solution:-

$\bf{= \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]-[ \sqrt{a}-x]} \times \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]+[ \sqrt{a}-x]}}$

$\bf \: = \frac{ [ \sqrt{a}+x]+[ \sqrt{a}-x]^{2} }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }$

$\bf \: = \frac{ [ \sqrt{a}+x]^{2}+[ \sqrt{a}-x]^{2}+2[ \sqrt{a}+x][ \sqrt{a}-x] }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }$

$\bf \: = \frac{4a}{4x \sqrt{a}}$

$\bf \: = \frac{ \sqrt{a} }{x}$

Now if

$\bf \: x=\frac{2ab}{1+ b^{2} }$

then we get

$\bf \: \frac{ \sqrt{a} }{2ab}*[1+ b^{2}]$

$\bf \: \frac{ [1+ b^{2}] }{2b \sqrt{a} }$