derviatives

Question

If f(x)=x^xln(2x−7), then f′(x)=x^xln(2x−7)g(x), where g(x) equals what?

Answer 1

I will assume that a pair of brackets was left out of your function, so your function is

 

f(x) = x^[x ln(2x-7)]

The general rule for the derivative of a function raised to a function is:

 

D_{x f} ^g = _f ^g[g'ln(f) + (g/f)f']                 (Equation 1)

Proof:
D_{x f} ^g = D_x e^(ln _f^g)                   (We simultaneously exponentiate and take the logarithm)
         = D_{x e}^{g(ln f)}                       (We apply a logarithm rule to bring down the g to the front)
         = _e^glnf(gln f)'                    (The derivative of e^f(x) is itself times the inside derivative)
         = _e^glnf[g' ln f + (g/f)f']      (Using the product rule and the chain rule)
         = _f ^g[g' ln f + (g/f)f']          (Finally, we undo the first two steps to the leading factor)

At this point, we notice that the quantity inside the brackets corresponds to the g(x) in your question.

Identifying the f and g in (Equation 1) as x and x ln(2x-7), respectively, we have

f = x

g = x ln(2x - 7)

f' = 1
g' = ln(2x - 7) + 2x/(2x - 7)

And therefore,
[g' ln f + (g/f)f'] = [ln(2x-7) + 2x/(2x-7)] ln x + [xln(2x-7)]/x·1

                        = [ln(2x-7) + 2x/(2x-7)] ln x + ln(2x-7)

Answer 2

f(x)=x^xln(2x−7) = h(x). w(x) where h(x) = x^x  and w(x) = ln (2x- 7)

You have to use the product rule to find the derivative of f(x).

First, find the derivative of h(x) = x^x

ln h(x) = ln x^x = x lnx 

Use implicit differentiation.

1/h(x) . h'(x) = ln x + x/x = ln x + 1

h'(x) = h(x) . (ln x + 1) = x^x. (ln x + 1)

 

The derivative of ln (2x-7) = (2) 1/(2x-7) = 2 / (2x-7)

 

f'(x) = h'(x). w(x) + h(x) . w'(x) = [x^x. (ln x + 1). ln (2x- 7)  + x^x . 2 / (2x-7)]The last step is to factor out the x^x ln(2x−7). This will leave you with the g(x) function. I leave this step for you to complete.