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!

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