How would you take the derivative of this function

Can someone explain to me how to do this process step by step?
I am using the product rule for this right?

Find the derivative of the function.
h(t) = (t^4 − 1)^7 (t^3 + 1)^4
h'(t) = ?

Do it step by step using the rules:
Let u(t) = t^4 − 1
v(t) = t^3 + 1
f(t) = ( u(t) )^ 7
g(t) = ( v(t) )^4
h(t) = f(t)*g(t)


h'(t) = f ' (t) * g(t) + f(t)* g ' (t) by the product rule

f ' (t) = ( 7*( u(t) )^ 6 ) * u ' (t) by exponent rule and chain rule
g ' (t) = ( 4* ( v(t) )^3 ) * v ' (t) by exponent rule and chain rule
u ' (t) = 4*t^3 by exponent rule
v ' (t) = 3* t^2 by exponent rule


Substitue u ' (t) and v ' (t) back into f ' (t) and g ' (t) to get
f ' (t) = ( 28*( u(t) )^ 6 ) * t^3
g ' (t) = ( 12* ( v(t) )^3 ) * t^2

Substitute f ' (t) and g ' (t) back into h ' (t) to get your final answer:
h'(t) = g(t) *( 28*( u(t) )^ 6 ) * t^3 ) + f(t)* ( 12* ( v(t) )^3 ) * t^2 )
= ( 28*( t^4 − 1 )^ 6 ) * t^3 * ( t^3 + 1 )^4 + ( 12* ( t^3 + 1 )^3 ) * t^2 *( t^4 − 1 )^7

Use Product Rule
h'(t) ={t^4-1)^7}{4*(t^3+1)^3 *(3t^2)} +{7*(t^4-1)^6*(4t^3)*(t^3+1)^4}
= [t^4-1)^6][(t^3+1)^3]{12t^2(t^4-1) +28t^3(t^3+1)}
=[t^4-1)^6][(t^3+1)^3]{12t^6 -12t^2) +28t^6+28t^3)}
=[t^4-1)^6][(t^3+1)^3]{40t^6 -12t^2 +28t^3)} ................Ans