That doesn't look like the chain rule to me. Apply the product rule first.
#3
jtt
16
0
i tried bringing down the 4Th exponent and then subtract it by one to get three, then leaving the inside alone ( 5x-1) at the same time taking the derivative of 3x^2. after that i got confused and got a wrong answer.
[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.