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, $\frac{dz}{d\theta}=-1.2\theta^{-\frac{5}{3}} + 0.88\theta^{-\frac{6}{5}}$;|undefined or, $\frac{dz}{d\theta}=\frac{0.88}{\sqrt[5]{\theta^6}}-\frac{1.2}{\sqrt[3]{\theta^5}}$.|undefined

(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.