How do you use the chain rule to differentiate #y=(4x^3-7)^4(3x+2)^10#?
1 Answer
I got:
#6(4x^3 - 7)^3(3x + 2)^9(44x^3 + 16x^2 - 35)#
If you can't validly modify your answer to match this one, you've mis-factored somewhere.
You would use the product rule on these terms, and the chain rule while using the product rule.
Product Rule:
#\mathbf(d/(dx)[f(x)g(x)] = f(x)g"'"(x) + g(x)f"'"(x))#
Chain Rule:
#\mathbf(d/(dx)[f(g(x))] = f"'"(g(x))*g"'"(x))#
Therefore, what you have (without the chain rule) is:
#d/(dx)[(4x^3 - 7)^4(3x + 2)^10]#
#= (4x^3 - 7)^4*10(3x + 2)^9 + (3x + 2)^10*4(4x^3 - 7)^3#
Now, incorporate the chain rule to finish this step and get:
#= (4x^3 - 7)^4*10(3x + 2)^9*3 + (3x + 2)^10*4(4x^3 - 7)^3*12x^2#
Now, group terms before factoring:
#= 30(4x^3 - 7)^4(3x + 2)^9 + 48x^2(4x^3 - 7)^3(3x + 2)^10#
Factor out common terms:
#= (4x^3 - 7)^3[30(4x^3 - 7)(3x + 2)^9 + 48x^2(3x + 2)^10]#
#= (4x^3 - 7)^3(3x + 2)^9[30(4x^3 - 7) + 48x^2(3x + 2)]#
Factor a little bit more:
#= 6(4x^3 - 7)^3(3x + 2)^9[5(4x^3 - 7) + 8x^2(3x + 2)]#
And maybe distribute inner terms:
#= 6(4x^3 - 7)^3(3x + 2)^9(20x^3 - 35 + 24x^3 + 16x^2)#
#= color(blue)(6(4x^3 - 7)^3(3x + 2)^9(44x^3 + 16x^2 - 35))#
That's probably simplified enough. Either way, it's correct.
