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derivativate (sqrt16-x)
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\[\sqrt{16-x}\]
is it (16-x)^2
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))
when there is a polynomial like x^n than the derivative will be: n*x^(n-1)
o ok cool thanks
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so here the sqrt mean 1/2 so sqrt x= x^1/2
ok thanks : )
no problem :-)
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