So, let's go ahead and plug in everything into the quotient rule:
For #f(x)/g(x)#: #d/dx(f(x)/g(x)) = (f'(x)*g(x) - g'(x)*f(x))/(g(x))^2#
So, all we need to do now is plug in #tan(x-7)# as #f(x)#, and #e^(2x) - 4# as #g(x)#. This gives us:
#(d/dx(tan(x-7))*(e^(2x) - 4) - d/dx(e^(2x) - 4)*tan(x-7))/(e^(2x) - 4)^2#
Now, the most tedious part is done. The last trick is to calculate the two derivatives.
Firstly, we need to find the derivative of #tan(x-7)#. The derivative of #tan(x)# is #sec^2(x)#. So, the derivative of #tan(x-7) = sec^2(x-7)#.
Note: you technically need to use the chain rule here, and multiply #sec^2(x-7)# by the derivative of #x-7#, but since the derivative of #x-7# is 1, we can ignore it.
Lastly, we need to find the derivative of #e^(2x)#. The derivative of #(e^(x)) = e^x#, but because of the #2x#, the chain rule comes into play. As a result, this would play out into:
#d/dx(e^(2x))*d/dx(2x) = e^(2x)*2 = 2e^(2x)#
So, plugging those in, we have the final answer:
#(sec^2(x-7)(e^(2x)-4)-2e^(2x)(tan(x-7)))/(e^(2x)-4)^2#
Now, you could distribute the #sec^2(x-7)#, or foil out the #(e^(2x)-4)^2#, but to be honest, you could just leave it as is.
Hope that helped :)
