Question

Using differentials, find the approximate values of the following:

(i) $\sqrt{25.02}$

(ii) (0.009)^1/3

(iii) (0.007)^1/3

(iv) $\sqrt{401}$

(v) (15)^1/4

(vi) (255)^1/4

(vii) $\frac{1}{(2.002{)}^2}$

(viii) log_e 4.04, it being given that log₁₀ 4 = 0.6021 and log₁₀ e = 0.4343.

(ix) log_e 10.02, it being given that log_e 10 = 2.3026.

(x) log₁₀ 10.1, it being given that log₁₀ e = 0.4343.

(xi) cos 61°, it being given that sin 60° = 0.86603 and 1° = 0.01745 radian.

(xii) $\frac{1}{\sqrt{25.1}}$

(xiii) $\sin \left(\frac{22}{14}\right)$

(xiv) $\cos \left(\frac{11\mathrm{\pi }}{36}\right)$

(xv) (80)^1/4

(xvi) (29)^1/3

(xvii) (66)^1/3

(xviii) $\sqrt{26}$

(xix) $\sqrt{37}$

(xx) $\sqrt{0.48}$

(xxi) (82)^1/4

(xxii) ${\left(\frac{17}{81}\right)}^{1/4}$

(xxiii) (33)^1/5

(xxiv) $\sqrt{36.6}$

(xxv) 25^1/3

Answer 1

(i)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt{x}. \\ \mathrm{Let}: \\ x=25 \\ x+∆x=25.02 \\ \mathrm{Then}, \\ ∆x=0.02 \\ \mathrm{For}x=25, \\ y=\sqrt{25}=5 \\ \mathrm{Let}: \\ dx=∆x=0.02 \\ \mathrm{Now},y=\sqrt{x} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=25}=\frac{1}{10} \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=\frac{1}{10}\times 0.02=0.002 \\ \Rightarrow ∆y=0.002 \\ \therefore \sqrt{25.02}=y+∆y=5.002 \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt[3]{x}. \\ \mathrm{Let}: \\ x=0.008 \\ x+∆x=0.009 \\ \mathrm{Then},∆x=0.001 \\ \mathrm{For}x=0.008, \\ y=\sqrt{0.008}=0.2 \\ \mathrm{Let}: \\ dx=∆x=0.001 \\ \mathrm{Now},y=\sqrt[3]{x} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0.008}=\frac{1}{3\times 0.04}=\frac{1}{0.12} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{0.12}\times 0.001=\frac{1}{120} \\ \Rightarrow ∆y=\frac{1}{120}=0.008333 \\ \therefore {\left(0.009\right)}^{\frac{1}{3}}=y+∆y=0.208333 \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt[3]{x.} \\ \mathrm{Let}: \\ x=0.008 \\ x+∆x=0.007 \\ \mathrm{Then},∆x=-0.001 \\ \mathrm{For}x=0.008, \\ y=\sqrt{0.008}=0.2 \\ \mathrm{Let}: \\ dx=∆x=-0.001 \\ \mathrm{Now},y=\sqrt[3]{x} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0.008}=\frac{1}{3\times 0.04}=\frac{1}{0.12} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{0.12}\times 0.001=\frac{1}{120} \\ \Rightarrow ∆y=\frac{1}{120}=0.008333 \\ \therefore {\left(0.007\right)}^{\frac{1}{3}}=y+∆y=0.191667 \end{gathered}$$

(iv).
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt{x}. \\ \mathrm{Let}: \\ x=400 \\ x+∆x=401 \\ \mathrm{Then}, \\ ∆x=1 \\ \mathrm{For}x=400, \\ y=\sqrt{400}=20 \\ \mathrm{Let}: \\ dx=∆x=1 \\ \mathrm{Now},y=\sqrt{x} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=400}=\frac{1}{40} \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=\frac{1}{40}\times 1=\frac{1}{40} \\ \Rightarrow ∆y=\frac{1}{40}=0.025 \\ \therefore \sqrt{401}=y+∆y=20.025 \end{gathered}$$

(v)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=x^{\frac{1}{4}}. \\ \mathrm{Let}: \\ x=16 \\ x+∆x=15 \\ \mathrm{Then}, \\ ∆x=-1 \\ \mathrm{For}x=16, \\ y={\left(16\right)}^{\frac{1}{4}}=2 \\ \mathrm{Let}: \\ dx=∆x=-1 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{4}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{4{\left(x\right)}^{{\displaystyle \frac{3}{4}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=16}=\frac{1}{32} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{32}\times \left(-1\right)=\frac{-1}{32} \\ \Rightarrow ∆y=\frac{-1}{32}=-0.03125 \\ \therefore {\left(15\right)}^{\frac{1}{4}}=y+∆y=1.96875 \end{gathered}$$

(vi)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{4}}. \\ \mathrm{Let}: \\ x=256 \\ x+∆x=255 \\ \mathrm{Then}, \\ ∆x=-1 \\ \mathrm{For}x=256, \\ y={\left(256\right)}^{\frac{1}{4}}=4 \\ \mathrm{Let}: \\ dx=∆x=-1 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{4}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{4{\left(x\right)}^{\frac{3}{4}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=256}=\frac{1}{256} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{256}\times -1=\frac{-1}{256} \\ \Rightarrow ∆y=\frac{-1}{256}=-0.003906 \\ \therefore {\left(255\right)}^{\frac{1}{4}}=y+∆y=3.99609\approx 3.9961 \end{gathered}$$

(vii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\frac{1}{x^2}. \\ \mathrm{Let}: \\ x=2 \\ x+∆x=2.002 \\ \mathrm{Then}, \\ ∆x=-0.002 \\ \mathrm{For}x=2, \\ y=\frac{1}{2^2}=\frac{1}{4} \\ \mathrm{Let}: \\ dx=∆x=0.002 \\ \mathrm{Now},y=\frac{1}{x^2} \\ \Rightarrow \frac{dy}{dx}=\frac{2}{x^3} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=2}=\frac{1}{4} \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=\frac{1}{4}\times -0.002=-0.0005 \\ \Rightarrow ∆y=-0.0005 \\ \therefore \frac{1}{{\left(2.002\right)}^2}=y+∆y=0.2495 \end{gathered}$$

(viii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\log }_{e}x. \\ \mathrm{Let}: \\ x=4 \\ x+∆x=4.04 \\ \mathrm{Then}, \\ ∆x=0.04 \\ \mathrm{For}x=4, \\ y={\log }_{e}4=\frac{{\log }_{10}4}{{\log }_{10}e}=\frac{0.6021}{0.4343}=1.386368 \\ \mathrm{Let}: \\ dx=∆x=0.04 \\ \mathrm{Now},y={\log }_{e}x \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=4}=\frac{1}{4} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{4}\times 0.04=0.01 \\ \Rightarrow ∆y=0.01 \\ \therefore {\log }_{e}4.04=y+∆y=1.396368 \end{gathered}$$

(ix)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\log }_{e}x. \\ \mathrm{Let}: \\ x=10 \\ x+∆x=10.02 \\ \mathrm{Then}, \\ ∆x=0.02 \\ \mathrm{For}x=, \\ y={\log }_{e}10=2.3026 \\ \mathrm{Let}: \\ dx=∆x=0.02 \\ \mathrm{Now},y={\log }_{e}x \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=10}=\frac{1}{10} \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=\frac{1}{10}\times 0.02=0.002 \\ \Rightarrow ∆y=0.002 \\ \therefore {\log }_{e}10.02=y+∆y=2.3046 \end{gathered}$$

(x)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\log }_{10}x. \\ \mathrm{Let}: \\ x=10 \\ x+∆x=10.1 \\ \mathrm{Then}, \\ ∆x=0.1 \\ \mathrm{For}x=, \\ y={\log }_{10}10=1 \\ \mathrm{Let}: \\ dx=∆x=0.1 \\ \mathrm{Now},y={\log }_{10}x=\frac{{\log }_{e}x}{{\log }_{e}10} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2.3025x} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=10}=0.04343 \\ \therefore ∆y=dy=\frac{dy}{dx}dx=0.04343\times 0.1=0.004343 \\ \Rightarrow ∆y=0.004343 \\ \therefore {\log }_{10}10.1=y+∆y=1.004343 \end{gathered}$$

(xi)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\cos x°. \\ \mathrm{Let}: \\ x=60° \\ x+∆x=61° \\ \mathrm{Then}, \\ ∆x=1°=0.01745 \\ \mathrm{For}x=60°, \\ y=\cos 60°=0.5 \\ \mathrm{Let}: \\ dx=∆x=0.01745 \\ \mathrm{Now},y=\cos x \\ \Rightarrow \frac{dy}{dx}=-\sin x \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=60}=-0.86603 \\ \therefore ∆y=dy=\frac{dy}{dx}dx=-0.86603\times 0.01745=-0.01511 \\ \Rightarrow ∆y=-0.01511 \\ \therefore \cos 61°=y+∆y=0.48488\approx 0.48489 \end{gathered}$$

(xii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\frac{1}{\sqrt{x}}. \\ \mathrm{Let}: \\ x=25 \\ x+∆x=25.1 \\ \mathrm{Then}, \\ ∆x=0.1 \\ \mathrm{For}x=, \\ y=\frac{1}{\sqrt{25}}=0.2 \\ \mathrm{Let}: \\ dx=∆x=0.1 \\ \mathrm{Now},y=\frac{1}{\sqrt{x}} \\ \Rightarrow \frac{dy}{dx}=\frac{-1}{2{\left(x\right)}^{{\displaystyle \frac{3}{2}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=25}=-0.004 \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=-0.004\times 0.1=-0.0004 \\ \Rightarrow ∆y=-0.0004 \\ \therefore \frac{1}{\sqrt{25.1}}=y+∆y=0.1996 \end{gathered}$$

(xiii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sin x. \\ \mathrm{Let}: \\ x=\frac{22}{7} \\ x+∆x=\frac{22}{14} \\ \mathrm{Then}, \\ ∆x=\frac{-22}{14} \\ \mathrm{For}x=\mathrm{\pi }, \\ y=\sin \left(\frac{22}{7}\right)=0 \\ \mathrm{Let}: \\ dx=∆x=\sin \frac{-22}{14}=-\sin \left(\frac{\mathrm{\pi }}{2}\right)=-1 \\ \mathrm{Now},y=\sin x \\ \Rightarrow \frac{dy}{dx}=\cos x \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{22}{7}}=-1 \\ \therefore ∆y=dy=\frac{dy}{dx}dx=-1\times \left(-1\right)=1 \\ \Rightarrow ∆y=1 \\ \therefore \sin \frac{22}{14}=y+∆y=1 \end{gathered}$$

(xiv)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\cos x. \\ \mathrm{Let}: \\ x=\frac{\mathrm{\pi }}{3} \\ x+∆x=\frac{11\mathrm{\pi }}{36} \\ \mathrm{Then}, \\ ∆x=\frac{-\mathrm{\pi }}{36}=-5° \\ \mathrm{For}x=\frac{\mathrm{\pi }}{3}, \\ y=\cos \left(\frac{\mathrm{\pi }}{3}\right)=0.5 \\ \mathrm{Let}: \\ dx=∆x=-\sin 5°=-0.08716 \\ \mathrm{Now},y=\cos x \\ \Rightarrow \frac{dy}{dx}=-\sin x \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{3}}=-0.86603 \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=-0.86603\times \left(-0.08716\right)=0.075575 \\ \Rightarrow ∆y=0.075575 \\ \therefore \cos \frac{11\mathrm{\pi }}{36}=y+∆y=0.5+0.075575=0.575575 \end{gathered}$$

(xv)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{4}}. \\ \mathrm{Let}: \\ x=81 \\ x+∆x=80 \\ \mathrm{Then}, \\ ∆x=-1 \\ \mathrm{For}x=81, \\ y={\left(81\right)}^{\frac{1}{4}}=3 \\ \mathrm{Let}: \\ dx=∆x=-1 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{4}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{4{\left(x\right)}^{{\displaystyle \frac{3}{4}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=81}=\frac{1}{108} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{108}\times \left(-1\right)=-0.009259 \\ \Rightarrow ∆y=-0.009259 \\ \therefore {\left(80\right)}^{\frac{1}{4}}=y+∆y=2.99074 \end{gathered}$$

(xvi)

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{3}}. \\ \mathrm{Let}: \\ x=27 \\ x+∆x=29 \\ \mathrm{Then}, \\ ∆x=2 \\ \mathrm{For}x=27, \\ y={\left(27\right)}^{\frac{1}{3}}=3 \\ \mathrm{Let}: \\ dx=∆x=2 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{3}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=27}=\frac{1}{27} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{27}\times 2=0.074 \\ \Rightarrow ∆y=0.074 \\ \therefore {\left(29\right)}^{\frac{1}{3}}=y+∆y=3.074 \end{gathered}$$

(xvii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{3}}. \\ \mathrm{Let}: \\ x=64 \\ x+∆x=66 \\ \mathrm{Then}, \\ ∆y=x=2 \\ \mathrm{For}x=64, \\ y={\left(64\right)}^{\frac{1}{3}}=4 \\ \mathrm{Let}: \\ dx=∆y=x=2 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{3}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=4}=\frac{1}{48} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{48}\times 2=0.04166 \\ \Rightarrow ∆y=0.04166 \\ \therefore {\left(66\right)}^{\frac{1}{3}}=y+∆y=4.0416 \end{gathered}$$

(xviii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt{x}. \\ \mathrm{Let}: \\ x=25 \\ x+∆x=26 \\ \mathrm{Then}, \\ ∆x=1 \\ \mathrm{For}x=25, \\ y=\sqrt{25}=5 \\ \mathrm{Let}: \\ dx=∆x=1 \\ \mathrm{Now},y={\left(x\right)}^{1/2} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=25}=\frac{1}{10} \\ \therefore ∆\mathrm{y}=dy=\frac{dy}{dx}dx=\frac{1}{10}\times 1=0.1 \\ \Rightarrow ∆y=0.1 \\ \therefore \sqrt{26}=y+∆y=5.1 \end{gathered}$$

(xix)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt{x}. \\ \mathrm{Let}: \\ x=36 \\ x+∆x=37 \\ \mathrm{Then}, \\ ∆x=1 \\ \mathrm{For}x=36, \\ y=\sqrt{36}=6 \\ \mathrm{Let}: \\ dx=∆x=1 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{2}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=36}=\frac{1}{12} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{12}\times 1=0.0833 \\ \Rightarrow ∆y=0.0833 \\ \therefore \sqrt{37}=y+∆y=6.0833 \end{gathered}$$

(xx)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt{x.} \\ \mathrm{Let}: \\ x=0.49 \\ x+∆x=0.48 \\ \mathrm{Then}, \\ ∆x=-0.01 \\ \mathrm{For}x=0.49, \\ y=\sqrt{0.49}=0.7 \\ \mathrm{Let}: \\ dx=∆x=0.01 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{2}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0.49}=\frac{1}{1.4} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{1.4}\times \left(-0.01\right)=-0.007143 \\ \Rightarrow ∆y=-0.007143 \\ \therefore \sqrt{0.48}=y+∆y=0.693 \end{gathered}$$

(xxi)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{4}}. \\ \mathrm{Let}: \\ x=81 \\ x+∆x=82 \\ \mathrm{Then}, \\ ∆x=1 \\ \mathrm{For}x=81, \\ y={\left(81\right)}^{\frac{1}{4}}=3 \\ Let: \\ dx=∆x=1 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{4}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{4{\left(x\right)}^{{\displaystyle \frac{3}{4}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=81}=\frac{1}{108} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{108}\times 1=0.009259 \\ \Rightarrow ∆y=0.009259 \\ \therefore {\left(82\right)}^{\frac{1}{4}}=y+∆y=3.009259 \end{gathered}$$

(xxii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{4}}. \\ \mathrm{Let}: \\ x=\frac{16}{81} \\ x+∆x=\frac{17}{81} \\ \mathrm{Then}, \\ ∆x=\frac{1}{81} \\ \mathrm{For}x=\frac{16}{81}, \\ y={\left(\frac{16}{81}\right)}^{\frac{1}{4}}=\frac{2}{3} \\ \mathrm{Let}: \\ dx=∆x=\frac{1}{81} \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{4}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{4{\left(x\right)}^{{\displaystyle \frac{3}{4}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{16}{81}}=\frac{27}{32} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{27}{32}\times \frac{1}{81}=\frac{1}{96}=0.01042 \\ \Rightarrow ∆y=0.01042 \\ \therefore {\left(\frac{17}{81}\right)}^{\frac{1}{4}}=y+∆y=0.6771 \end{gathered}$$

(xxiii)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{5}}. \\ \mathrm{Let}: \\ x=32 \\ x+∆x=33 \\ \mathrm{Then}, \\ ∆x=1 \\ \mathrm{For}x=33, \\ y={\left(32\right)}^{\frac{1}{5}}=2 \\ \mathrm{Let}: \\ dx=∆x=1 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{5}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{5{\left(x\right)}^{{\displaystyle \frac{4}{5}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=32}=\frac{1}{80} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{80}\times 1=0.0125 \\ \Rightarrow ∆y=0.0125 \\ \therefore {\left(33\right)}^{\frac{1}{5}}=y+∆y=2.0125 \\ \mathrm{Disclaimer}:\mathrm{This}\mathrm{solution}\mathrm{has}\mathrm{been}\mathrm{created}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}.\mathrm{However},\mathrm{the}\mathrm{solution}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}\mathrm{is}\mathrm{incorrect}. \end{gathered}$$

(xxiv)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)=\sqrt{x}. \\ \mathrm{Let}: \\ x=36 \\ x+∆x=36.6 \\ \mathrm{Then}, \\ ∆x=0.6 \\ \mathrm{For}x=36, \\ y=\sqrt{36}=6 \\ \mathrm{Let}: \\ dx=∆x=0.6 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{2}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=36}=\frac{1}{12} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{12}\times 0.6=0.05 \\ \Rightarrow ∆y=0.05 \\ \therefore \sqrt{36.6}=y+∆y=6.05 \end{gathered}$$

(xv)
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}y=f\left(x\right)={\left(x\right)}^{\frac{1}{3}}. \\ \mathrm{Let}: \\ x=27 \\ x+∆x=25 \\ \mathrm{Then}, \\ △x=-2 \\ \mathrm{For}x=27, \\ y={\left(27\right)}^{\frac{1}{3}}=3 \\ \mathrm{Let}: \\ dx=∆x=-2 \\ \mathrm{Now},y={\left(x\right)}^{\frac{1}{3}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=27}=\frac{1}{27} \\ \therefore ∆y=dy=\frac{dy}{dx}dx=\frac{1}{27}\times \left(-2\right)=-0.07407 \\ \Rightarrow ∆y=-0.07407 \\ \therefore {\left(25\right)}^{\frac{1}{3}}=y+∆y=2.9259 \end{gathered}$$