Let f, g : (a, b)→R be differentiable functions and h(x) = f (x)g(x).
Then h being product differentiable functions, is differentiable on (a,b) 
So by definition dh = h'(x)dx.
Now by using product rule we have h'(x) = f (x)g'(x) + f'(x)g(x).
Thus dh = h'(x)dx = ( f (x)g'(x) + f''(x)g(x))dx = f (x)g'(x)dx + f '(x)g(x) dx
= f (x)dg + g(x)df = fdg + gdf

Let x be the number
Let y = f(x) = \(x^\frac{1}{n}\)
Then log y = \(\frac1n\) log x
Taking differential on both sides we get,
\(\frac { 1 }{ y } dy=\frac { 1 }{ n } \times \frac { 1 }{ x } dx\)
i.e. \(\frac { \Delta y }{ y } \simeq \frac { dy }{ y } =\frac { 1 }{ n } .\frac { dx }{ x } \)
\(\therefore \frac { \Delta y }{ y } \times 100\simeq \frac { 1 }{ n } \left( \frac { dx }{ x } \times 100 \right) \)
\(\simeq \frac { 1 }{ n } \) times the percentage error in the number. Hence, percentage error in the nth root of a number is approximately \(\frac1n\) times the percentage error in the number

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Volume of sphere = \(\frac43\)πr²
Given r = 10 cm 
\(\frac{dr}{dt}\) = - 0.2
V = \(\frac43\)πr³
Change in Volume
= \(\frac{4}{\not 3} \pi . \not 3 r^{2} \frac{d r}{d t}\)
= 4π(10)² (-0.2)
= 400 π (-0.2) = -80 πcm³
∴ Volume decreases by 80 π cm³
Surface area of sphere = 4πr²
Change 10 surrace area = 4 π2r\(\frac{dr}{dt}\)
= 8π(10) (-0.2)
= -\(\frac{80π\times2}{10}\) = -16 π cm²
∴ Surface area decreases by 16 π cm²

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18