roselin
find the derivative of u=5x+1/2 sqrt of x
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Lexxiie123
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what is sqrt
Lexxiie123
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is it square root?
roselin
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yes
mark_o.
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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
mark_o.
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d(kx)/dx=k.,,dx^n/dx= nx^n-1
roselin
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the problem is actually: |dw:1351567010524:dw|
mark_o.
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u=5x + 1/(2sqrtx)?
roselin
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\[u= 5x+1\div 2\sqrt{x}\]
mark_o.
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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
roselin
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what if i use the quotient rule?
mark_o.
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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 ) ?
roselin
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its find the derivative of the function \[u= 5x+1 \div \sqrt{2}\]
mark_o.
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ok the quotient rule says
\[\frac{ d(\frac{ U }{ V }) }{ dx }=\frac{ V \frac{ du }{ dx }-U \frac{ dV }{ dx } }{ v ^{2} }\]
roselin
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\[u= 5x+1\div 2\sqrt{x}\]
roselin
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so i must use the power rule?
mark_o.
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arrange them like
\[u=5x +(\frac{ 1 }{ 2 })(\frac{ 1 }{ x ^{\frac{ 1 }{ 2 }} })\]
roselin
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okay
mark_o.
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what did you get ?
roselin
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give me a minute
mark_o.
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lets try to work only on
1/x^1/2
mark_o.
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u=1, v=x^1/2
roselin
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where did u get the v from?
mark_o.
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if you are using the quotient rule
Dx(U/V)
roselin
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wait the quotient rule is |dw:1351568697070:dw|
roselin
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that is the quotient rule right?
mark_o.
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lets try to work only on
1/x^1/2
therefore
Dx(U/V)=Dx( 1/x^1/2) so here u=1 and v=x^1/2
roselin
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its 2
roselin
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sorry,
mark_o.
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yes on youR formula g=V..,f=u , its the same..you guys have taught f and g but variables such u,v, w are used also locally and internationally
roselin
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okay,
mark_o.
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ok try to use your formula and you will arrive with the same answer anyway
roselin
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i am solving it here, give me a minute i will post what i have
mark_o.
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ok take your time..lol.. :D
roselin
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\[-5x+6/ 2\sqrt{x}\]
roselin
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i got that answer
roselin
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u thre?
mark_o.
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hm try to study this one
let u=1 and v= x^1/2
Dx(1/(x)^1/2)
x^1/2(Dx(1) - 1 Dx(x^1/2)
= ---------------------------
( x^1/2)^2
0 - (1/2)x^-1/2
= ----------------
x
-(1/2)
= ---------
x x^1/2
-(1/2)
=--------------
x^3/2
-1
= ----------
2 x^3/2
mark_o.
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now lets go back to the prob
derivative of u=5x+(1/2)( x)^1/2
1
du/dx= 5 - ------------
2( 2x^3/2)
1
= 5- -------------
4x^3/2
w/c is the same as doing it w/ power rule