How do I differentiate the function of dy/dx?

Well, I'm really not sure why you are confused.

Though, dy/dx (6*ln(x)) = 6/x

a probability derivation: d/dx(y) = d/dx(f(x)^(g(x))) The by-product of y is 0: 0 = d/dx(f(x)^(g(x))) tutor f(x)^(g(x)) as a a probability of e: f(x)^(g(x)) = e^(log(f(x)^(g(x)))) = e^(g(x) log(f(x))): 0 = d/dx(e^(g(x) log(f(x)))) utilising the chain rule, d/dx(e^(g(x) log(f(x)))) = ( de^u)/( du) ( du)/( dx), the position u = g(x) log(f(x)) and ( d)/( du)(e^u) = e^u: 0 = (d/dx(g(x) log(f(x)))) e^(g(x) log(f(x))) tutor e^(g(x) log(f(x))) as a a probability of f(x): e^(g(x) log(f(x))) = e^(log(f(x)^(g(x)))) = f(x)^(g(x)): 0 = f(x)^(g(x)) d/dx(g(x) log(f(x))) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), the position u = g(x) and v = log(f(x)): 0 = log(f(x)) (d/dx(g(x)))+g(x) (d/dx(log(f(x)))) f(x)^(g(x)) The by-product of g(x) is g'(x): 0 = f(x)^(g(x)) (g(x) (d/dx(log(f(x))))+g'(x) log(f(x))) utilising the chain rule, d/dx(log(f(x))) = ( dlog(u))/( du) ( du)/( dx), the position u = f(x) and ( d)/( du)(log(u)) = a million/u: 0 = f(x)^(g(x)) ((d/dx(f(x)))/(f(x)) g(x)+log(f(x)) g'(x)) The by-product of f(x) is f'(x): answer: | | 0 = f(x)^(g(x)) ((f'(x) g(x))/(f(x))+log(f(x)) g'(x)) it is the way you ensure your different question a probability derivation: d/dx(y) = d/dx(f(x) g(x) h(x)) The by-product of y is 0: 0 = d/dx(f(x) g(x) h(x)) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), the position u = f(x) and v = g(x) h(x): 0 = g(x) h(x) (d/dx(f(x)))+f(x) (d/dx(g(x) h(x))) The by-product of f(x) is f'(x): 0 = f(x) (d/dx(g(x) h(x)))+f'(x) g(x) h(x) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), the position u = g(x) and v = h(x): 0 = h(x) (d/dx(g(x)))+g(x) (d/dx(h(x))) f(x)+g(x) h(x) f'(x) The by-product of g(x) is g'(x): 0 = f(x) (g(x) (d/dx(h(x)))+g'(x) h(x))+g(x) h(x) f'(x) The by-product of h(x) is h'(x): answer: | | 0 = g(x) h(x) f'(x)+f(x) (h'(x) g(x)+h(x) g'(x))