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Maxiney=5-2x
y'(u-v)=du/dx-dv/dx
y'(5-2x)=d/dx(5)-d/dx(2x)
-The derivative of 5 is 0 because it's a constant.
-The derivative of 2x is:
d/dx(cu)=c*du/dx where c is a constant.
d/dx(2x)=2*d/dx(x)
y'(5-2x)=(0)-[2*d/dx(x)]
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
y'(5-2x)=-[2*(1)]
y'(5-2x)=-(2)
y'(5-2x)= -2
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