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OpenStudy (anonymous):
what is sqrt
OpenStudy (anonymous):
is it square root?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
derivative of u=5x+(1/2) sqrt of x
sometimes its easier to use the power rule, and rewriting it as
u=5x+(1/2 )( x)^1/2 use power rule of the derivative
OpenStudy (anonymous):
d(kx)/dx=k.,,dx^n/dx= nx^n-1
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OpenStudy (anonymous):
the problem is actually: |dw:1351567010524:dw|
OpenStudy (anonymous):
u=5x + 1/(2sqrtx)?
OpenStudy (anonymous):
\[u= 5x+1\div 2\sqrt{x}\]
OpenStudy (anonymous):
it will be the same idea by using the power rule
u=5x + 1/(2sqrtx)
u=5x + (1/2)(x)^-1/2 now find the derivative
OpenStudy (anonymous):
what if i use the quotient rule?
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OpenStudy (anonymous):
u=5x + (1/2)(x)^-1/2
is it
du/dt=5+(1/2)(-1/2)x^-1/2 - 2/2)
du/dt=5+(1/2)(-1/2)x^-3/2 ) ?
OpenStudy (anonymous):
its find the derivative of the function \[u= 5x+1 \div \sqrt{2}\]
OpenStudy (anonymous):
ok the quotient rule says
\[\frac{ d(\frac{ U }{ V }) }{ dx }=\frac{ V \frac{ du }{ dx }-U \frac{ dV }{ dx } }{ v ^{2} }\]
OpenStudy (anonymous):
\[u= 5x+1\div 2\sqrt{x}\]
OpenStudy (anonymous):
so i must use the power rule?
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OpenStudy (anonymous):
arrange them like
\[u=5x +(\frac{ 1 }{ 2 })(\frac{ 1 }{ x ^{\frac{ 1 }{ 2 }} })\]
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
what did you get ?
OpenStudy (anonymous):
give me a minute
OpenStudy (anonymous):
lets try to work only on
1/x^1/2
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