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OpenStudy (anonymous):

derivativate (sqrt16-x)

OpenStudy (anonymous):

\[\sqrt{16-x}\]

OpenStudy (anonymous):

is it (16-x)^2

OpenStudy (anonymous):

first you use the chain rule so take the derivative of 16-x, that is just -1 than multiply -1 with the derivative of the sqrt that is 1/2 * (16-x)^(-1/2) so the solution is: - 1/2 (16-x)^(-1/2)= -1/ (2 sqrt(16-x))

OpenStudy (anonymous):

when there is a polynomial like x^n than the derivative will be: n*x^(n-1)

OpenStudy (anonymous):

o ok cool thanks

OpenStudy (anonymous):

so here the sqrt mean 1/2 so sqrt x= x^1/2

OpenStudy (anonymous):

ok thanks : )

OpenStudy (anonymous):

no problem :-)

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