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Mathematics 12 Online
OpenStudy (anonymous):

Not sure why I forgot but I can't seem to figure out how to find the derivative of this. Can anyone help? f[x]= C/(x-1).... I know the answer I just need the steps to refresh my memory. thanks

OpenStudy (anonymous):

do you recall the derivative of 1/x?

OpenStudy (anonymous):

-1/x^(-2)

OpenStudy (anonymous):

| Factor out constants: = | c (d/dx(1/(x-1))) | Use the chain rule, d/dx(1/(x-1)) = d/( du)1/u ( du)/( dx), where u = x-1 and d/( du)1/u = -1/u^2: = | c (-(d/dx(x-1))/(x-1)^2) | Differentiate the sum term by term: = | -(c (d/dx(x)+d/dx(-1)))/(x-1)^2 | The derivative of -1 is zero: = | -(c (d/dx(x)+0))/(x-1)^2 | The derivative of x is 1: = | (c -1 × 1)/(x-1)^2

OpenStudy (anonymous):

f[x]= C/(x-1). = c(x-1)^-1 f'(x)= -c(x-1)^-2 = -c/(x-1)^2

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