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Shanon Bosco

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โˆ™ 2021-02-27 00:47:19
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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โˆ™ 2011-11-23 05:57:31

ln(1)/[1-x]?

d/dx(u/v)=(v*du/dx-u*dv/dx)/(v2)

d/dx(ln(1)/[1-x])=[(1-x)*d/dx(ln1)-ln1*d/dx(1-x)]/[(1-x)2]

-The derivative of ln1 is:

d/dx(lnu)=(1/u)*d/dx(u)

d/dx(ln1)=(1/1)*d/dx(1)

d/dx(ln1)=(1)*d/dx(1)

d/dx(ln1)=d/dx(1)

-The derivative of 1-x is:

d/dx(u-v)=du/dx-dv/dx

d/dx(1-x)=d/dx(1)-d/dx(x)

d/dx(ln(1)/[1-x])=[(1-x)*d/dx(1)-ln1*(d/dx(1)-d/dx(x))]/[(1-x)2]

-The derivative of 1 is 0 because it is a constant.

-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

d/dx(ln(1)/[1-x])=[(1-x)*(0)-ln1*(0-1)]/[(1-x)2]

d/dx(ln(1)/[1-x])=[-ln1*(-1)]/[(1-x)2]

d/dx(ln(1)/[1-x])=[ln1]/[(1-x)2]

But you see, ln1 is equal to 0:

d/dx(ln(1)/[1-x])=[0]/[(1-x)2]

d/dx(ln(1)/[1-x])=0

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