Question

Examine the function
f(x) = (xe^[(.5)(x-2)^2])*(x+2)^3
Use logarithmic differentiation to calculate the derivative of f(x) . At what values does logarithmic
differentiation fail?

Answer 1

\[f(x)=(xe^{(.5)(x-2)^2})*(x+2)^3 \]
\[ln(f(x))=ln((xe^{(.5)(x-2)^2})*(x+2)^3)=ln(x) + ln(e^{(.5)(x-2)^2}) + ln((x+2)^3)\]
\[=ln(x) + (.5)(x-2)^2 + 3*ln(x+2)\]
Apply the rules of logarithms to this function.
\[ln(\frac{x}{y})=ln(x) - ln(y)\]
\[ln(x*y)=ln(x) + ln(y)\]
\[ln(x^n)=n*ln(x)\]
\[ln(e^(x^n))=x^n\]

You should be able to differentiate more easily now.

Answer 2

OH! Okay, I was going to say this is ridiculous.

Answer 3

Thank you

Answer 4

this is your answer , enjoy http://mathway.com/answer.aspx?p=calg?p=f(x)SMB15SMB01SMB15(xeSMB07SMB08(.5)(x-2)SMB072SMB09)*(x+2)SMB073?p=132?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=EvaluateSMB15theSMB15Function

Answer 5

maybe

Answer 6

What would be the best method of finding the tangent line of this ugly thing?

Answer 7

A tangent line is basically in the form of y=mx + b
to put it in better variables, t(a) = f'(a)*x + (f(a)-f'(a)*a)
Where your y-intercept (b=f(a)-f'(a)*a)

Answer 8

right but finding where f '(x) = 0 would be terrible