Differentiation by the chain rule

That doesn't look like the chain rule to me. Apply the product rule first.

 

[itex] y = f(x)g(x)[/itex] where [itex]f(x)= x^3[/itex] and [itex]g(x)=(5x-1)^4[/itex]
So you'll first need to apply the product rule... as you do you'll need the derivative of g.

[itex] g(x) = P\circ L (x) = P( L(x))[/itex] where [itex] P(x) = x^4[/itex] and [itex] L(x)=5x - 1[/itex]. As a composition you need to apply the chain rule. (P for power, L for linear).

If you'd rather use the Leibniz notation form of the chain rule: [itex]\frac{du}{dx} = \frac{du}{dv} \frac{dv}{dx}[/itex] then let u=g(x) = P(v) with v = L(x).

 

Your function is f*g where f=x^3 and g=(5x-1)^4, right? The product rule says the derivative of f*g is f'*g+f*g', also right? Now you just need to find f' and g'. Finding the derivative of g' is where you need the chain rule.