We'll have to use a little more than the chain rule to get our result.
#x^3# and #(7x-7)^5# are being multiplied by each other, so we have to use the product rule, which states
#d/dx(uv)=u'v+uv'#
Where #u# and #v# are functions of #x#. Let's have #u=x^3# and #v=(7x-7)^5#. Now we differentiate them:
#u=x^3->u'=3x^2#
#v=(7x-7)^5->v'=5*7(7x-7)^4=35(7x-7)^4#
In differentiating #v#, we have to use the chain rule - meaning we find the derivative of the "inside" function #(7x-7)# and multiply it by the "outside" function #((7x-7)^5)#. The derivative of #7x-7# is #7#, and #7# multiplied by #(7x-7)^5# is #7(7x-7)^5#. To finish off, use the power rule to get #35(7x-7)^4#.
Plugging everything we just found into the product rule formula, we find
#y'=(x^3)'(7x-7)^5+(x^3)((7x-7)^5)'#
#y'=3x^2(7x-7)^5+35x^3(7x-7)^4#
If we want to, we can factor out an #x^2(7x-7)^4# to simplify the result to:
#y'=x^2(7x-7)^4(3(7x-7)+35x)#
#y'=x^2(7x-7)^4(56x-21)#
We can go even further to get:
#y'=(56x^3-21x^2)(7x-7)^4#
