If you'd like additional help on the concepts associated with this problem, I have some videos that can help:
Product & Quotient Rules
Practice Problems
Anyways, onto the problem:
For this, you'll actually need a product rule in addition to the quotient rule, because of the product you have in your numerator #(xtanx)#. But that's some way off. The first step for this problem is the quotient rule:
The formula for the quotient rule is:
#d/dx[color(blue)(f(x))/color(red)(g(x))] = (color(blue)(f'(x))*color(red)(g(x)) - color(red)(g'(x))*color(blue)(f(x)))/[color(red)g(x)]^2#
So, we know that #f(x) = xtan(x)# and #g(x) = x+4#. Now we just plug this into the equation:
#f'(x) = (color(blue)(d/dx(xtanx))*color(red)(x+4) - color(red)(d/dx(x+4))*color(blue)(xtanx))/[color(red)(x+4)]^2#
And now, we just take a few derivatives.
#d/dx(x+4)# is very straightforward -- the derivative of any linear function is it's slope, and #x+4# has a slope of 1, so:
#d/dx(x+4) = 1#
Now, #d/dx(xtanx)# is a wee bit tricker because, as we mentioned before, you'll need to invoke the product rule to take this derivative.
This is as follows:
#d/dx(color(blue)(f(x))**color(green)(g(x))) = color(blue)(f'(x))color(green)(g(x)) + color(green)(g'(x))color(blue)(f(x))#
We know that #f(x) = x# and #g(x) = tan(x)# (It doesn't matter which you pick to be which so long as you are consistent). So we just plug in:
#=> d/dx(color(blue)(tan(x))**color(green)(x)) = color(blue)(d/dx(tanx))color(green)((x)) + color(green)(d/dx(x))color(blue)(tan(x))#
#= (color(blue)sec^2(x)*color(green)x) + (color(green)1*color(blue)tan(x))#
#= xsec^2(x) + tan(x)#
Now we plug this back into our quotient rule. This leaves us with:
#f'(x) = (color(blue)[[xsec^2(x) + tan(x)]]*(color(red)(x+4)) -[ color(red)[(1)*color(blue)(xtanx)]])/[color(red)(x+4)]^2#
Some quick simplification:
#f'(x) = [(x+4)[xsec^2(x) + tan(x)] - xtan(x)]/(x+4)^2#
And that's your final answer.
Hope that helped :)
