anonymous
  • anonymous
find the derivative of u=5x+1/2 sqrt of x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
what is sqrt
anonymous
  • anonymous
is it square root?
anonymous
  • anonymous
yes

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anonymous
  • 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
anonymous
  • anonymous
d(kx)/dx=k.,,dx^n/dx= nx^n-1
anonymous
  • anonymous
the problem is actually: |dw:1351567010524:dw|
anonymous
  • anonymous
u=5x + 1/(2sqrtx)?
anonymous
  • anonymous
\[u= 5x+1\div 2\sqrt{x}\]
anonymous
  • 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
anonymous
  • anonymous
what if i use the quotient rule?
anonymous
  • 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 ) ?
anonymous
  • anonymous
its find the derivative of the function \[u= 5x+1 \div \sqrt{2}\]
anonymous
  • anonymous
ok the quotient rule says \[\frac{ d(\frac{ U }{ V }) }{ dx }=\frac{ V \frac{ du }{ dx }-U \frac{ dV }{ dx } }{ v ^{2} }\]
anonymous
  • anonymous
\[u= 5x+1\div 2\sqrt{x}\]
anonymous
  • anonymous
so i must use the power rule?
anonymous
  • anonymous
arrange them like \[u=5x +(\frac{ 1 }{ 2 })(\frac{ 1 }{ x ^{\frac{ 1 }{ 2 }} })\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
what did you get ?
anonymous
  • anonymous
give me a minute
anonymous
  • anonymous
lets try to work only on 1/x^1/2
anonymous
  • anonymous
u=1, v=x^1/2
anonymous
  • anonymous
where did u get the v from?
anonymous
  • anonymous
if you are using the quotient rule Dx(U/V)
anonymous
  • anonymous
wait the quotient rule is |dw:1351568697070:dw|
anonymous
  • anonymous
that is the quotient rule right?
anonymous
  • anonymous
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
anonymous
  • anonymous
its 2
anonymous
  • anonymous
sorry,
anonymous
  • anonymous
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
anonymous
  • anonymous
okay,
anonymous
  • anonymous
ok try to use your formula and you will arrive with the same answer anyway
anonymous
  • anonymous
i am solving it here, give me a minute i will post what i have
anonymous
  • anonymous
ok take your time..lol.. :D
anonymous
  • anonymous
\[-5x+6/ 2\sqrt{x}\]
anonymous
  • anonymous
i got that answer
anonymous
  • anonymous
u thre?
anonymous
  • anonymous
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
anonymous
  • anonymous
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

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