ANSWERS.
Exercises I. (p. 25.)
(1) d y d x = 13 x 12 {\displaystyle {\dfrac {dy}{dx}}=13x^{12}} .
(2) d y d x = − 3 2 x − 5 2 {\displaystyle {\dfrac {dy}{dx}}=-{\dfrac {3}{2}}x^{-{\frac {5}{2}}}} .
(3) d y d x = 2 a x ( 2 a − 1 ) {\displaystyle {\dfrac {dy}{dx}}=2ax^{(2a-1)}} .
(4) d u d t = 2.4 t 1.4 {\displaystyle {\dfrac {du}{dt}}=2.4t^{1.4}} .
(5) d z d u = 1 3 u − 2 3 {\displaystyle {\dfrac {dz}{du}}={\dfrac {1}{3}}u^{-{\frac {2}{3}}}} .
(6) d y d x = − 5 3 x − 8 3 {\displaystyle {\dfrac {dy}{dx}}=-{\dfrac {5}{3}}x^{-{\frac {8}{3}}}} .
(7) d u d x = − 8 5 x − 13 5 {\displaystyle {\dfrac {du}{dx}}=-{\dfrac {8}{5}}x^{-{\frac {13}{5}}}} .
(8) d y d x = 2 a x a − 1 {\displaystyle {\dfrac {dy}{dx}}=2ax^{a-1}} .
(9) d y d x = 3 q x 3 − q q {\displaystyle {\dfrac {dy}{dx}}={\dfrac {3}{q}}x^{\frac {3-q}{q}}} .
(10) d y d x = − m n x − m + n n {\displaystyle {\dfrac {dy}{dx}}=-{\dfrac {m}{n}}x^{-{\frac {m+n}{n}}}} .
Exercises II. (p. 33.)
(1) d y d x = 3 a x 2 {\displaystyle {\dfrac {dy}{dx}}=3ax^{2}} .
(2) d y d x = 13 × 3 2 x 1 2 {\displaystyle {\dfrac {dy}{dx}}=13\times {\frac {3}{2}}x^{\frac {1}{2}}} .
(3) d y d x = 6 x − 1 2 {\displaystyle {\dfrac {dy}{dx}}=6x^{-{\frac {1}{2}}}} .
(4) d y d x = 1 2 c 1 2 x − 1 2 {\displaystyle {\dfrac {dy}{dx}}={\dfrac {1}{2}}c^{\frac {1}{2}}x^{-{\frac {1}{2}}}} .
(5) d u d z = a n c z n − 1 {\displaystyle {\dfrac {du}{dz}}={\dfrac {an}{c}}z^{n-1}} .
(6) d y d t = 2.36 t {\displaystyle {\dfrac {dy}{dt}}=2.36t} .
(7) d l t d t = 0.000012 × l 0 {\displaystyle {\dfrac {dl_{t}}{dt}}=0.000012\times l_{0}} .
(8) d C d V = a b V b − 1 {\displaystyle {\dfrac {dC}{dV}}=abV^{b-1}} , 0.98 {\displaystyle 0.98} , 3.00 {\displaystyle 3.00} and 7.47 {\displaystyle 7.47} candle power per volt respectively.
(9) d n d D = − 1 L D 2 g T π σ , d n d L = − 1 D L 2 g T π σ {\displaystyle {\dfrac {dn}{dD}}=-{\dfrac {1}{LD^{2}}}{\sqrt {\dfrac {gT}{\pi \sigma }}},{\dfrac {dn}{dL}}=-{\dfrac {1}{DL^{2}}}{\sqrt {\dfrac {gT}{\pi \sigma }}}} . d n d σ = − 1 2 D L g T π σ 3 , d n d T = 1 2 D L g π σ T {\displaystyle \;\quad {\dfrac {dn}{d\sigma }}=-{\dfrac {1}{2DL}}{\sqrt {\dfrac {gT}{\pi \sigma ^{3}}}},{\dfrac {dn}{dT}}={\dfrac {1}{2DL}}{\sqrt {\dfrac {g}{\pi \sigma T}}}} .