How do you differentiate #f(x)=sec(1/(5x^2-1))# using the chain rule?

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Answer 1 · 2016-06-24T18:51:19

Let #y = secu#

and #u = (5x^2 - 1)^(-1)#

Now, we have to differentiate #y# and #u#. We will have to use the chain rule to differentiate u.

#u = v^-1#

#v = 5x^2 - 1#

#d/dx((5x^2 - 1)^-1) = 10x xx -1/(v^(2))#

#=-(10x)/v^2#

#= -(10x)/(5x^2 - 1)^2#

Hence, #u' = -(10x)/(5x^2 - 1)^2#.

The derivative of #secu# is #-cscu#, because #secu = 1/cosu# and #(1/cosu)' = -1/sinu = -cscu#.

#f'(x) = -cscu xx -(10x)/(5x^2 - 1)^2#

#f'(x) = (10xcsc(1/(5x^2 - 1)))/(5x^2 - 1)^2#

Hopefully this helps!