Let f ( x ) = ( ln x ) 2 , f ' (e^4)

Answer 1

Let f ( x ) = ( ln x ) 2 , f ' (e⁴) = ??

Detailed Solution:

I am going to assume, it is: f ( x ) = ( ln x )²

f ^' ( x ) = (2) ( ln x ) * ( 1/x ) <== Chain Rule

f ^' ( x ) = ( 2 ln x ) / (x )

f ^' ( e⁴ ) = ( 2 ln (e⁴) ) / (e⁴ )

f ^' ( e⁴ ) = ( 2*4 ln (e) ) / (e⁴ )

f ^' ( e⁴ ) = ( 2*4*(1) ) / (e⁴ )

f ^' ( e⁴ ) = 8 / (e⁴ ) = 8e^-4 <== Exact Solution

Answer 2

I'm assuming here that you were looking for

f(x)=(ln(x))².

To take the derivative, I'll rewrite this so that it looks nicer and we can use the product rule

f(x)=(ln(x))(ln(x))

To restate, the product rule for taking the derivative of a function which is the product of two functions, f*g, is

if y = f*g,

then y'=f' * g + g' * f

f=ln(x)

f'=1/x

g=ln(x)

g'=1/x

so f * g' = ln(x)/x, f' * g = ln(x)/x, (f * g') + (f' * g) = ln(x)/x + ln(x)/x = 2ln(x)/x

So, f'(x)=2ln(x)/x

Now, to evaluate f(e⁴), just plug in

f(e⁴)=2ln(e⁴)/e⁴

=8ln(e)/e⁴

We know ln(e)=1, so this then becomes 8/e⁴, which can also be written as 8*e^(-4).

I hope this makes sense! Let me know if I can help you with anything else, good luck!