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Alexandra [31]
3 years ago
7

What is the derivative of x to the minus 5 power

Mathematics
2 answers:
Leni [432]3 years ago
7 0
(x^{-5})'=-5x^{-6}
Rudiy273 years ago
3 0
Using implicit differentiation:

y={ x }^{ -5 }\\ \\ y=\frac { 1 }{ { x }^{ 5 } } \\ \\ { x }^{ 5 }\cdot y=\frac { 1 }{ { x }^{ 5 } } \cdot { x }^{ 5 }

\\ \\ { x }^{ 5 }\cdot y=1\\ \\ { x }^{ 5 }\frac { dy }{ dx } +5{ x }^{ 4 }\cdot y=0

\\ \\ { x }^{ 5 }\frac { dy }{ dx } =-5{ x }^{ 4 }\cdot y\\ \\ { x }^{ 5 }\frac { dy }{ dx } =-5{ x }^{ 4 }\cdot \frac { 1 }{ { x }^{ 5 } }

\\ \\ \frac { 1 }{ { x }^{ 5 } } \cdot { x }^{ 5 }\frac { dy }{ dx } =-5{ x }^{ 4 }\cdot \frac { 1 }{ { x }^{ 5 } } \cdot \frac { 1 }{ { x }^{ 5 } }

\\ \\ \frac { dy }{ dx } =-\frac { 5{ x }^{ 4 } }{ { x }^{ 10 } } \\ \\ =-\frac { 5 }{ { x }^{ 6 } }
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Recall your d = rt, distance = rate * time

now, if say, by the time they meet, Mr Cunningham has travelled "d" miles, that means Mrs Cunningham must also had travelled "d" miles as well.

However, he left 3 hours earlier, so by the time he travelled "d" miles, and took say "t" hours, for her it took 3 hour less, because she started driving 3 hours later, so, she's been on the road 3 hours less than Mr Cunningham, so by the time they meet, Mrs Cunningham has travelled then "t - 3" hours.

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{Mr Cunningham}&d&40&t\\
\textit{Mrs Cunningham}&d&80&t-3
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=40t\\
d=80(t-3)\\
----------\\
\boxed{40t}=80(t-3)
\end{cases}
\\\\\\
\cfrac{40t}{80}=t-3\implies \cfrac{t}{2}=t-3\implies t=2t-6\implies 6=2t-t
\\\\\\
6=t
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