Find d/dx ((3a-5)x^a)/(5a-4)

I assume you mean d/dx.With this being a quotient, we would use the quotient rule. With that, the top is always the hardest.In general, if we are given f(x)/g(x), the quotient rule states:d/dx of f(x)/g(x) = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)*2]Following this, we would need the derivative of the top and the bottom. So:d/dx (3a-5)x^a = a(3a-5)x^(a-1)d/dx (5a-4) = 0 (when it's with respect to x, it's like "a" is a constant. So, this would be like taking the derivative of "6.")So, combining everything:d/dx (3a-5)x^a/(5a-4) = [(5a-4)a(3a-5)x^(a-1) - (3a-5)x^a*0]/(5a-4)^2 = (5a-4)a(3a-5)x^(a-1)/(5a-4)^2And, then, you can factor out a (5a-4), for. . . = a(3a-5)x^(a-1)/(5a-4)Which would be the same as if you took everything with the "a's" in the parenthesis as one factor. As in. . .
(3a-5)/(5a-4) ---> K, so ((3a-5)x^a)/(5a-4) ---> Kx^aThe d/dx of this is. . .a*K*x^(a-1)Replace the K with (3a-5)/(5a-4), we get the same answer.