Differentiate e^x/9+9x

Differentiate the function (find y')
y= (e^x)/9+9x

Using quotient rule, d(u/v)/dx = (vu' - uv')/ v^2

Then, d(e^x/9+9x) = ((9+9x)e^x - e^x (9))/(9+9x)^2
= 9xe^x / (9+9x)^2
= xe^x / 9 (1+x)^2

Remember the quotient rule. If h(x) = k(x) / m(x), then:

h'(x) = [m(x)k'(x) - k(x)m'(x)] / [m(x)]²

So, here (I presume, though your notation is ambiguous):

k(x) = e^x
m(x) = (9 + 9x)

h'(x) = [(9 + 9x)(e^x) - (e^x)(9)] / [9 + 9x]²
h'(x) = [9e^x + 9xe^x - 9e^x] / [9 + 9x]²
h'(x) = 9xe^x / (9 + 9x)²
h'(x) = xe^x / (1 + x)²

y= (e^x)/9+9x
y'=(e^x)/9 + 9