(3) y = ϵ 2 x x + 1 {\displaystyle y=\epsilon ^{\frac {2x}{x+1}}} .
log ϵ y = 2 x x + 1 , 1 y d y d x = 2 ( x + 1 ) − 2 x ( x + 1 ) 2 ; h e n c e d y d x = 2 ( x + 1 ) 2 ϵ 2 x x + 1 {\displaystyle {\begin{aligned}\log _{\epsilon }y&={\frac {2x}{x+1}},\quad {\frac {1}{y}}\,{\frac {dy}{dx}}={\frac {2(x+1)-2x}{(x+1)^{2}}};\\hence{\frac {dy}{dx}}&={\frac {2}{(x+1)^{2}}}\epsilon ^{\frac {2x}{x+1}}\end{aligned}}} .
Check by writing 2 x x + 1 = z {\displaystyle {\dfrac {2x}{x+1}}=z} .
(4) y = ϵ x 2 + a {\displaystyle y=\epsilon ^{\sqrt {x^{2}+a}}} . log ϵ y = ( x 2 + a ) 1 2 {\displaystyle \log _{\epsilon }y=(x^{2}+a)^{\frac {1}{2}}} .
1 y d y d x = x ( x 2 + a ) 1 2 and d y d x = x × ϵ x 2 + a ( x 2 + a ) 1 2 {\displaystyle {\frac {1}{y}}\,{\frac {dy}{dx}}={\frac {x}{(x^{2}+a)^{\frac {1}{2}}}}\quad {\text{and}}\quad {\frac {dy}{dx}}={\frac {x\times \epsilon ^{\sqrt {x^{2}+a}}}{(x^{2}+a)^{\frac {1}{2}}}}} .
(For if ( x 2 + a ) 1 2 = u {\displaystyle (x^{2}+a)^{\frac {1}{2}}=u} and x 2 + a = v {\displaystyle x^{2}+a=v} , u = v 1 2 {\displaystyle u=v^{\frac {1}{2}}} ,
d u d v = 1 2 v 1 2 ; d v d x = 2 x ; d u d x = x ( x 2 + a ) 1 2 ) {\displaystyle {\frac {du}{dv}}={\frac {1}{{2v}^{\frac {1}{2}}}};\quad {\frac {dv}{dx}}=2x;\quad {\frac {du}{dx}}={\frac {x}{(x^{2}+a)^{\frac {1}{2}}}})} .
Check by writing x 2 + a = z {\displaystyle {\sqrt {x^{2}+a}}=z} .
(5) y = log ( a + x 3 ) {\displaystyle y=\log(a+x^{3})} . Let ( a + x 3 ) = z {\displaystyle (a+x^{3})=z} ; then y = log ϵ z {\displaystyle y=\log _{\epsilon }z} .
d y d z = 1 z ; d z d x = 3 x 2 ; hence d y d x = 3 x 2 a + x 3 {\displaystyle {\frac {dy}{dz}}={\frac {1}{z}};\quad {\frac {dz}{dx}}=3x^{2};\quad {\text{hence}}\quad {\frac {dy}{dx}}={\frac {3x^{2}}{a+x^{3}}}} .
(6) y = log ϵ { 3 x 2 + a + x 2 } {\displaystyle y=\log _{\epsilon }\{{3x^{2}+{\sqrt {a+x^{2}}}}\}} . Let 3 x 2 + a + x 2 = z {\displaystyle 3x^{2}+{\sqrt {a+x^{2}}}=z} ; then y = log ϵ z {\displaystyle y=\log _{\epsilon }z} .
d y d z = 1 z ; d z d x = 6 x + x x 2 + a ; d y d x = 6 x + x x 2 + a 3 x 2 + a + x 2 = x ( 1 + 6 x 2 + a ) ( 3 x 2 + x 2 + a ) x 2 + a . {\displaystyle {\begin{aligned}{\frac {dy}{dz}}&={\frac {1}{z}};\quad {\frac {dz}{dx}}=6x+{\frac {x}{\sqrt {x^{2}+a}}};\\{\frac {dy}{dx}}&={\frac {6x+{\dfrac {x}{\sqrt {x^{2}+a}}}}{3x^{2}+{\sqrt {a+x^{2}}}}}={\frac {x(1+6{\sqrt {x^{2}+a}})}{(3x^{2}+{\sqrt {x^{2}+a}}){\sqrt {x^{2}+a}}}}.\end{aligned}}} .