Page:Calculus Made Easy.pdf/171

(3) $y=\epsilon ^{\frac {2x}{x+1}}$.

${\begin{aligned}\log _{\epsilon }y&={\frac {2x}{x+1}},\quad {\frac {1}{y}}\,{\frac {dy}{dx}}={\frac {2(x+1)-2x}{(x+1)^{2}}};\\hence{\frac {dy}{dx}}&={\frac {2}{(x+1)^{2}}}\epsilon ^{\frac {2x}{x+1}}\end{aligned}}$.

Check by writing ${\dfrac {2x}{x+1}}=z$.

(4) $y=\epsilon ^{\sqrt {x^{2}+a}}$. $\log _{\epsilon }y=(x^{2}+a)^{\frac {1}{2}}$.

${\frac {1}{y}}\,{\frac {dy}{dx}}={\frac {x}{(x^{2}+a)^{\frac {1}{2}}}}\quad {\text{and}}\quad {\frac {dy}{dx}}={\frac {x\times \epsilon ^{\sqrt {x^{2}+a}}}{(x^{2}+a)^{\frac {1}{2}}}}$.

(For if $(x^{2}+a)^{\frac {1}{2}}=u$ and $x^{2}+a=v$, $u=v^{\frac {1}{2}}$,

${\frac {du}{dv}}={\frac {1}{{2v}^{\frac {1}{2}}}};\quad {\frac {dv}{dx}}=2x;\quad {\frac {du}{dx}}={\frac {x}{(x^{2}+a)^{\frac {1}{2}}}})$.

Check by writing ${\sqrt {x^{2}+a}}=z$.

(5) $y=\log(a+x^{3})$. Let $(a+x^{3})=z$; then $y=\log _{\epsilon }z$.

${\frac {dy}{dz}}={\frac {1}{z}};\quad {\frac {dz}{dx}}=3x^{2};\quad {\text{hence}}\quad {\frac {dy}{dx}}={\frac {3x^{2}}{a+x^{3}}}$.

(6) $y=\log _{\epsilon }\{{3x^{2}+{\sqrt {a+x^{2}}}}\}$. Let $3x^{2}+{\sqrt {a+x^{2}}}=z$; then $y=\log _{\epsilon }z$.

${\begin{aligned}{\frac {dy}{dz}}&={\frac {1}{z}};\quad {\frac {dz}{dx}}=6x+{\frac {x}{\sqrt {x^{2}+a}}};\\{\frac {dy}{dx}}&={\frac {6x+{\dfrac {x}{\sqrt {x^{2}+a}}}}{3x^{2}+{\sqrt {a+x^{2}}}}}={\frac {x(1+6{\sqrt {x^{2}+a}})}{(3x^{2}+{\sqrt {x^{2}+a}}){\sqrt {x^{2}+a}}}}.\end{aligned}}$.