How do you differentiate # f(x)=e^((lnx-2)^2 # using the chain rule.?

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Answer 1 · 2016-07-11T14:37:02

This will require two applications of the chain rule.

Let's start with what we have in the exponent.

Let #y = u^2# and #u = lnx - 2#

The derivative of #y# is #2u# and the derivative of #u# is #1/x - 0 = 1/x#.

Hence, #dy/dx = 2u xx 1/x = (2u)/x = (2(lnx - 2))/x#

Now, let #y = e^u# and #u = (2(lnx - 2))/x#.

We already know the derivative of #u#. As for #y#, the derivative of #e^u# is #e^u#.

So, #f'(x) = e^((2(lnx - 2))/x)#.

Hopefully this helps!