How do you differentiate #f(x)=1/(16x+3)^2# using the chain rule.?

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Answer 1 · 2015-12-31T01:27:09

#f'(x)=(-32)/(16x+3)^3#

Write as:

#f(x)=(16x+3)^-2#

Now, use the power rule with the chain rule:

#d/dx(u^-2)=-2u^-3#, and #u=16x+3#.

Thus,

#f'(x)=-2(16x+3)^-3d/dx(16x+3)#

#=>(-2)/(16x+3)^-3 * 16#

#=>(-32)/(16x+3)^3#