How do I find the derivative of #y = ln (9x)/(1+x)#?

Question

Ezekiel Alexander · Accepted Answer

The quotient rule states that, given an expression #f(x) = g(x)/(h(x))#, the derivative #f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^2#

In the case above, our functions are:

#g(x) = ln(9x)# and #h(x) = 1+x#.

Giving us derivatives of:

#g'(x) = (d/dx)ln(9x) = 9/(9x) = 1/x# and #h'(x) = (d/dx)(1+x) = 1#

Thus, by the Quotient Rule:

#f'(x)= ((1/x)(1+x) - (ln(9x))(1))/(1+x)^2 = (1/x + 1 - ln (9x))/(x^2+2x+1) #

Note that #g(x)# is undefined in the real number plane for any #x<=0#; therefore, the derivative will be undefined as well (recall that #ln(x)# does not have a real value for #x<=0#). Thus, since the only point at which our denominator is 0 would be #x=-1#, we need not worry about the denominator causing discontinuities within the domain of our function #f(x)#.

Christopher Allers · Answer

undefined To find the derivative of $ y = \ln\left(\frac{9x}{1+x}\right) $, you can use the chain rule and the quotient rule. The derivative is:

$$ y' = \frac{d}{dx}\left[\ln\left(\frac{9x}{1+x}\right)\right] = \frac{1}{\frac{9x}{1+x}} \cdot \frac{d}{dx}\left[\frac{9x}{1+x}\right] $$

$$ = \frac{1}{\frac{9x}{1+x}} \cdot \left[\frac{9(1+x) - 9x}{(1+x)^2}\right] $$

$$ = \frac{1}{\frac{9x}{1+x}} \cdot \left[\frac{9}{1+x}\right] $$

$$ = \frac{1}{\frac{9x}{1+x}} \cdot \frac{9}{1+x} $$

$$ = \frac{1}{x} \cdot \frac{9}{1+x} $$

$$ = \frac{9}{x(1+x)} $$