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question
- Thread starter Nathan23
- Start date
Hello, Nathan23!
Secondly, you can't even multiply it out correctly! \(\displaystyle \;\;(3x^4\,+\,8)(3x^4\,+\,8)\;=\;9x^8\,+\,48x\,+\,64\)
First of all, they don't want you to multiply it out . . . they want you to use the Product Rule.Use the product rule to find the derivative: \(\displaystyle f(x)\:=\: (3x^4\,+\,8)^2\)
I am getting \(\displaystyle f'(x)\:=\:9x^{16}\,+\, 64\;\;\) Am I messing this up? \(\displaystyle \;\) . . . you sure are!
Secondly, you can't even multiply it out correctly! \(\displaystyle \;\;(3x^4\,+\,8)(3x^4\,+\,8)\;=\;9x^8\,+\,48x\,+\,64\)
so
they want you to find the derivative of f(x)= (3x^4 +8)(3x^4 +8)
so f'(x)= (3x^4 + 8)(12x^3) + (3x^4 + 8)(12x^3)
so f'(x)= (36x^7 + 96x^3) + (36x^7 + 96x^3)
so f'(x)= 72x^7 + 192x^3 so f'(x)=24(3x^7 + 8x^3)
that is product rule which is if
f(x) = g(x)h(x)
then
f'(x) = g(x)h'(x) + h(x)g'(x)
an easier way is called the chain rule which for this problem is
f(x)= (3x^4 + 8)^2
f'(x)= 2(3x^4 + 8)(12x^3)
f'(x)= 2(36x^7 + 96x^3) or
f'(x)=24(3x^7 + 8x^3)
Does that explain it?
Peace,
KC
they want you to find the derivative of f(x)= (3x^4 +8)(3x^4 +8)
so f'(x)= (3x^4 + 8)(12x^3) + (3x^4 + 8)(12x^3)
so f'(x)= (36x^7 + 96x^3) + (36x^7 + 96x^3)
so f'(x)= 72x^7 + 192x^3 so f'(x)=24(3x^7 + 8x^3)
that is product rule which is if
f(x) = g(x)h(x)
then
f'(x) = g(x)h'(x) + h(x)g'(x)
an easier way is called the chain rule which for this problem is
f(x)= (3x^4 + 8)^2
f'(x)= 2(3x^4 + 8)(12x^3)
f'(x)= 2(36x^7 + 96x^3) or
f'(x)=24(3x^7 + 8x^3)
Does that explain it?
Peace,
KC