How do you use the chain rule to differentiate #1/-(4x)#?

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Answer 1 · 2016-09-16T17:44:35

#(1) / (4 x^(2))#

We have: #(1) / (- (4 x))#

This expression can be differentiated using the "chain rule".

Let #u = - 4 x => u' = - 4# and #v = (1) / (u) = u^(- 1) => v' = - 1 u^(- 2) = - (1) / (u^(2))#:

#=> (d) / (dx) ((1) / (- (4 x))) = - 4 cdot - (1) / (u^(2))#

#=> (d) / (dx) ((1) / (- (4 x))) = (4) / (u^(2))#

We can now replace #u# with #- 4 x#:

#=> (d) / (dx) ((1) / (- (4 x))) = (4) / ((- 4 x)^(2))#

#=> (d) / (dx) ((1) / (- (4 x))) = (4) / (16 x^(2))#

#=> (d) / (dx) ((1) / (- (4 x))) = (1) / (4 x^(2))#