help!! differentiate wrt x: e^-cx^2*cos(logx)?

Question

CALCULUS PEOPLE!

Answer 1

Expression to differtiate

e^-cx^2cos(log x)

Let y= e^-cx^2cos(log x)

y= e^{-cx^[2cos(log x)]}

Take log on both sides to base e

log y = log { e^{-cx^[2cos(log x)]}}
= {-cx^[2cos(log x)]} * log e

{ because log a^ [exp] = [exp] * log a}

log y = {-cx^[2cos(log x)]}
(-1/c)*log y = x^[2cos(log x)]

let d = -1/c constant
d*log y = x^[2cos(log x)]

Again take log on both sides to base e

log (d *log y) = log {x^([2cos(log x)])}
log d+log(log y) = 2cos(log x) *log x

Now differentiate w.r.to x

1/(log y) * (1/y) * dy/dx = 2cos(log x) * 1/x + log x * (-2sin(log x) * (1/x) )

(1/(y*log y) *dy/dx = (1/x) { 2cos (log x) - 2 (log x) * sin(log x) }

dy/dx = y*(log y) * (1/x) * { 2cos (log x) - 2 (log x) * sin(log x) }

dy/dx = e^{-cx^[2cos(log x)]} * {-cx^[2cos(log x)]} * (1/x) * { 2cos (log x) - 2 (log x) * sin(log x) }



I hope i remember maths i've not been in touch with maths for years

Answer 2

yeah,u need to put in more paranthesis to clear whether -cx^2 is with -c or with e...??!!

Answer 3

The question is ambiguous. Please edit the question by using more parentheses to make it clear.