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Mathematics 7 Online
OpenStudy (anonymous):

find the derivative of u=5x+1/2 sqrt of x

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

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?

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?

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

OpenStudy (anonymous):

u=1, v=x^1/2

OpenStudy (anonymous):

where did u get the v from?

OpenStudy (anonymous):

if you are using the quotient rule Dx(U/V)

OpenStudy (anonymous):

wait the quotient rule is |dw:1351568697070:dw|

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