Using differentials, find the approximate value of , [tex]\tt \bigg(0.999 \bigg)^{ \dfrac{1}{10 } }[/tex] upto three places of decimal.
[tex] \rule{300pt}{0.1pt}[/tex]
Answer with proper explanation.
Don't Dare for spam.
→ Expecting answer from :
★ Moderators
★ Brainly stars and teacher
★ Others best users
[tex] \rule{300pt}{0.1pt}[/tex]
Answer with proper explanation.
Don't Dare for spam.
→ Expecting answer from :
★ Moderators
★ Brainly stars and teacher
★ Others best users
Answers & Comments
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that
[tex]\sf \: f(x) = \bigg(x \bigg)^{ \dfrac{1}{10 } }, \: \: \: where \: x = 1 \\ \\ [/tex]
So,
[tex]\sf \: f(x + \triangle x\:) = \bigg(x + \triangle x\: \bigg)^{ \dfrac{1}{10 } }, \: \: \: where \: \triangle x\: = - 0.001 \\ \\ [/tex]
Now, using definition of differentials, we have
[tex]\sf \: f(x + \triangle x\:) = f(x) + f'(x)\triangle x\: \\ \\ [/tex]
So, on substituting the values, we get
[tex]\sf \: \bigg(x + \triangle x\:\bigg)^{ \dfrac{1}{10 } } = \bigg(x \bigg)^{ \dfrac{1}{10 } } + \dfrac{1}{10} {\bigg(x \bigg) }^{ - \dfrac{9}{10} } \: \triangle x\: \\ \\ [/tex]
So, on substituting the values, we get
[tex]\sf \: \bigg(1 - 0.001\:\bigg)^{ \dfrac{1}{10 } } = \bigg(1\bigg)^{ \dfrac{1}{10 } } + \dfrac{1}{10} {\bigg(1\bigg) }^{ - \dfrac{9}{10} } \:( - 0.001)\: \\ \\ [/tex]
[tex]\sf \: \bigg(0.999\:\bigg)^{ \dfrac{1}{10 } } = 1 - \dfrac{0.001}{10} \\ \\ [/tex]
[tex]\sf \: \bigg(0.999\:\bigg)^{ \dfrac{1}{10 } } = 1 - 0.0001 \\ \\ [/tex]
[tex]\bf\implies \: \bigg(0.999\:\bigg)^{ \dfrac{1}{10 } } = 0.9999 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]
Solution:-
Let assume that
[tex]\begin{gathered}\sf \: f(x) = \bigg(x \bigg)^{ \dfrac{1}{10 } }, \: \: \: where \: x = 1 \\ \\ \end{gathered}
[tex]\begin{gathered}\sf \: f(x) = \bigg(x \bigg)^{ \dfrac{1}{10 } }, \: \: \: where \: x = 1 \\ \\ \end{gathered} [/tex]
So,
[tex]f(x+△x)=(x+△x)101,where△x=−0.001
[/tex]
Now, using definition of differentials, we have
[tex]\begin{gathered}\sf \: f(x + \triangle x\:) = f(x) + f'(x)\triangle x\: \\ \\ \end{gathered}[/tex]
So, on substituting the values, we get
[tex]\begin{gathered}\sf \: \bigg(x + \triangle x\:\bigg)^{ \dfrac{1}{10 } } = \bigg(x \bigg)^{ \dfrac{1}{10 } } + \dfrac{1}{10} {\bigg(x \bigg) }^{ - \dfrac{9}{10} } \: \triangle x\: \\ \\ \end{gathered}
[tex]\begin{gathered}\sf \: \bigg(x + \triangle x\:\bigg)^{ \dfrac{1}{10 } } = \bigg(x \bigg)^{ \dfrac{1}{10 } } + \dfrac{1}{10} {\bigg(x \bigg) }^{ - \dfrac{9}{10} } \: \triangle x\: \\ \\ \end{gathered} [/tex]
So, on substituting the values, we get
[tex]\begin{gathered}\sf \: \bigg(1 - 0.001\:\bigg)^{ \dfrac{1}{10 } } = \bigg(1\bigg)^{ \dfrac{1}{10 } } + \dfrac{1}{10} {\bigg(1\bigg) }^{ - \dfrac{9}{10} } \:( - 0.001)\: \\ \\ \end{gathered}
[tex]\begin{gathered}\sf \: \bigg(1 - 0.001\:\bigg)^{ \dfrac{1}{10 } } = \bigg(1\bigg)^{ \dfrac{1}{10 } } + \dfrac{1}{10} {\bigg(1\bigg) }^{ - \dfrac{9}{10} } \:( - 0.001)\: \\ \\ \end{gathered} [/tex]
[tex](0.999) ^{ \frac{1}{10} } =1−0.0001
1[/tex]
[tex]⟹(0.999)^{ \frac{1}{10} } =0.9999
[tex]⟹(0.999)^{ \frac{1}{10} } =0.9999[/tex]