(i) \( y =\left(x^{2}+5\right)^{3}, x=1, d x=0.05 \)
\(d y =3\left(x^{2}+5\right)^{2} \times 2 x d x \)
\( =3(1+5)^{2} \times 2 \times 0.05 \)
\( =3 \times 36 \times 0.1 \)
dy = 10.8
(ii) \( y =\cos x, x=\pi / 6, d x=0.05 \)
\(d y =-\sin x d x \)
\( =-\sin \pi / 6 \times 0.05 \)
\( =-\frac{1}{2} \times 0.05 \)
dy = -0.025

(i) \( \mathrm{y} =\left(x^{4}+x^{2}+1\right)^{1 / 2} \)
\(\mathrm{dy} =\frac{1}{2}\left(x^{4}+x^{2}+1\right)^{-1 / 2}\left(4 x^{3}+2 x\right) d x \)
\( =\frac{2\left(2 x^{3}+x\right)}{2\left(x^{4}+x^{2}+1\right)^{1 / 2}} d x=\frac{\left(2 x^{3}+x\right)}{\left(x^{4}+x^{2}+1\right)^{1 / 2}} d x\)
(ii) \( \mathrm{y} =\left(x^{4}+x^{2}+1\right)^{1 / 2} \)
\(\mathrm{dy} =\frac{1}{2}\left(x^{4}+x^{2}+1\right)^{-1 / 2}\left(4 x^{3}+2 x\right) d x \)
\( =\frac{2\left(2 x^{3}+x\right)}{2\left(x^{4}+x^{2}+1\right)^{1 / 2}} d x=\frac{\left(2 x^{3}+x\right)}{\left(x^{4}+x^{2}+1\right)^{1 / 2}} d x \)

The volume of the cube of side is x
\( V =x^{3} \)
\(d V =3 x^{2} d x \)
\(d V =3 x^{2} \times(0.02 x) \)
\( =0.06 x^{3} \mathrm{~m}^{3}\)

Area of a circular template A =  \(\pi\)r²
dA = 2\(\pi\)rdr
\(\mathrm{dA}=2 \times \pi \times 10 \times 0.02\)
\(=0.4 \pi \mathrm{cm}^{2}\)
The possible error in calculating the template area is approximately 0.4\(\pi\) cm².

Volume of the sphere V \(=\frac{4}{3} \pi r^{3}\)
\( d V=\frac{4}{3} \pi 3 r^{2} d r \)
\( d V=4 \pi(21)^{2}(0.05) \)
= 277
The maximum error in the calculated volume is about 277 cm³.

We have f(2) = 2³ + 2² - 2(2) + 1
= 9
\(f(2.05)=(2.05)^{3}+(2.05)^{2}-2(2.05)+1\)
= 9.717625
\(\Delta y=f(x+\Delta x)-f(x)\)
= 9.717625 - 9
= 0.717 625
dy = f'(x)dx
= (3x²+2x-2)dx
= (12+4-2) \(\times\) 0.005
= 0.7

\( f(x) =x^{3}+2 x^{2} \)
\(f^{\prime}(x) =3 x^{2}+4 x \)
\(d y =\left(3 x^{2}+4 x\right) d x \)
\(d y =(3(4)+4(2)) 0.1 \)
 = (12+8) 0.1 
\( =20 \times 0.1=2\)

1.1

33x – 150

% error in P = 0.7%

\(sin\left( \frac { 22 }{ 14 } \right) =1\)

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } ={ e }^{ xy }\left[ 3xy+1+{ x }^{ 2 }{ y }^{ 2 } \right] \)

\(\frac { dw }{ d\theta } =0\)

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } \)

\(\frac { dw }{ dt } ={ 6a }^{ 2 }{ t }^{ 2 }e^{ 2{ a }^{ 2 } }{ t }^{ 3 }\)