Page:Calculus Made Easy.pdf/62

(3) Differentiate $z=1.8{\sqrt[{3}]{\dfrac {1}{\theta ^{2}}}}-{\dfrac {4.4}{\sqrt[{5}]{\theta }}}-27^{\circ }$.

This may be written: $z=1.8\theta ^{-{\frac {2}{3}}}-4.4\theta ^{-{\frac {1}{5}}}-27^{\circ }$

The $27^{\circ }$ vanishes, and we have

${\frac {dz}{d\theta }}=1.8\times -{\tfrac {2}{3}}\times \theta ^{-{\frac {2}{3}}-1}-4.4\times \left(-{\tfrac {1}{5}}\right)\theta ^{-{\frac {1}{5}}-1}$;

or,

${\frac {dz}{d\theta }}=-1.2\theta ^{-{\frac {5}{3}}}+0.88\theta ^{-{\frac {6}{5}}}$;

or,

${\frac {dz}{d\theta }}={\frac {0.88}{\sqrt[{5}]{\theta ^{6}}}}-{\frac {1.2}{\sqrt[{3}]{\theta ^{5}}}}$.

(4) Differentiate $v=(3t^{2}-1.2t+1)^{3}$ A direct way of doing this will be explained later (see p.67); but we can nevertheless manage it now without any difficulty.

Developing the cube, we get

$v=27t^{6}-32.4t^{5}+39.96t^{4}-23.328t^{3}+13.32t^{2}-3.6t+1$;

hence

${\frac {dv}{dt}}=162t^{5}-162t^{4}+159.84t^{3}-69.984t^{2}+26.64t-3.6$.

(5) Differentiate $y=(2x-3)(x+1)^{2}$

${\frac {dy}{dx}}=(2x-3){\frac {d\left[(x+1)(x+1)\right]}{dx}}+(x+1)^{2}{\frac {d(2x-3)}{dx}}$

$=(2x-3)\left[(x+1){\frac {d(x+1)}{dx}}\right.+\left.(x+1){\frac {d(x+1)}{dx}}\right]$

$+(x+1)^{2}{\frac {d(2x-3)}{dx}}$

$=2(x+1)\left[(2x-3)+(x+1)\right]=2(x+1)(3x-2)$;

or, more simply, multiply out and then differentiate.