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How do you differentiate #f(x)=arctan(1/(2x-1))# using the chain rule?

1 Answer

mason m · Stefan V.

Nov 23, 2015

#f'(x)=-1/(2x^2-2x+1)#

Explanation:

According the Chain Rule, #d/dx[arctan(u)]=(u')/(1+u^2)#

If you want explanation of how to find this, please ask.

Therefore, #f'(x)=overbrace(d/dx[1/(2x-1)])^(color(blue)("write as " (2x-1)^(-1)))/(1+(1/(2x-1))^2#.

Simplify and get: #f'(x)=-1/(2x^2-2x+1)#

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