(9) Differentiate y = x 2 x 2 + 1 {\displaystyle y={\frac {x^{2}}{x^{2}+1}}} .
d y d x = ( x 2 + 1 ) 2 x − x 2 × 2 x ( x 2 + 1 ) 2 = 2 x ( x 2 + 1 ) 2 {\displaystyle {\frac {dy}{dx}}={\frac {(x^{2}+1)\,2x-x^{2}\times 2x}{(x^{2}+1)^{2}}}={\frac {2x}{(x^{2}+1)^{2}}}} .
(10) Differentiate
y = a + x a − x {\displaystyle y={\frac {a+{\sqrt {x}}}{a-{\sqrt {x}}}}} .
In the indexed form,
y = a + x 1 2 a − x 1 2 {\displaystyle y={\frac {a+x^{\frac {1}{2}}}{a-x^{\frac {1}{2}}}}} .
d y d x = ( a − x 1 2 ) ( 1 2 x − 1 2 ) − ( a + x 1 2 ) ( − 1 2 x − 1 2 ) ( a − x 1 2 ) 2 = a − x 1 2 + a + x 1 2 2 ( a − x 1 2 ) 2 x 1 2 {\displaystyle {\frac {dy}{dx}}={\frac {(a-x^{\frac {1}{2}})({\tfrac {1}{2}}x^{-{\frac {1}{2}}})-(a+x^{\frac {1}{2}})(-{\tfrac {1}{2}}x^{-{\frac {1}{2}}})}{(a-x^{\frac {1}{2}})^{2}}}={\frac {a-x^{\frac {1}{2}}+a+x^{\frac {1}{2}}}{2(a-x^{\frac {1}{2}})^{2}x^{\frac {1}{2}}}}} ;
hence
d y d x = a ( a − x ) 2 x {\displaystyle {\frac {dy}{dx}}={\frac {a}{(a-{\sqrt {x}})^{2}\,{\sqrt {x}}}}} .
(11) Differentiate
θ = 1 − a t 2 3 1 + a t 3 2 {\displaystyle \theta ={\frac {1-a{\sqrt[{3}]{t^{2}}}}{1+a{\sqrt[{2}]{t^{3}}}}}} .
Now
θ = 1 − a t 2 3 1 + a t 3 2 {\displaystyle \theta ={\frac {1-at^{\frac {2}{3}}}{1+at^{\frac {3}{2}}}}} .
(12) A reservoir of square cross-section has sides sloping at an angle of 45 ∘ {\displaystyle 45^{\circ }} with the vertical. The side