how do i derivate this function?

Question

d [5e^x]
___ ______
dx [4+9e^x]





d [2x^7/5 - 8]
___ ______
dx [ e^x + 5]




d Cosx + x^3
___ ______
dx 7tanx




d
___ √tan(3x) (square root of all tan(3x) )
dx




d(e^x cosx (x^2 -1))
________________
dx

Answer 1

d/dx f(x)/g(x) = f'(x)/g(x) - f(x)*g'(x)/[g(x)]^2

so for the first, f(x) = 5e^x, g(x) = 4 + 9e^x

f'(x) = 5e^x g'(x) = 9e^x

d/dx f(x)/g(x) = [5e^x]/[4+9e^x] - [5e^x*9e^x] / [4+9e^x]^2

d/dx f(x)/g(x) = [5e^x]/[4+9e^x] - [45e^2x] / [4+9e^x]^2

Use the same method for the second and third parts. For the next,

d/dx √[f(x)] = 0.5*1/√[f(x)] * df(x)/dx

d/dx [tan(ax)] = a*sec^2(ax)

d/dx √[tan(3x)] = 0.5/√[tan(3x)] * 3*sec^2(3x) = [1.5*sec^2(3x)] / √[tan(3x)]

d/dx [e^x * cosx * (x^2 - 1)] use the product rule twice:

d/dx [e^x * cosx] * [x^2 - 1] = d/dx[e^x * cosx] * [x^2 - 1] + [2x] * [e^x * cosx] =

[-e^x*sinx + e^x*cosx] * [x^2 - 1] + [2x] * [e^x * cosx]

you can multiply it out and collect terms to simplify