If f(x)= (x^2+3x+7)^4, find F'(x)?

Question

I can not seem to find the right answer!!!

Answer 1

This involves chain rule, meaning that the derivative of a function composition is equal to the derivative of the outer function multiplied by the derivative of the inner. For this problem, derive the outer fuction, x^4, and multiply by the inner, x^2 + 3x + 7.
= 4(x^2 + 3x + 7)^3 * (2x +3)
= (8x + 12)(x^2 + 3x + 7)^3 (combining the 4 and the 2x +3)

Answer 2

f ' (x) = 4 ( x ² + 3 x + 7 ) ³ ( 2 x + 3 )

Answer 3

calculus
well you have the power rule
4(x^2+3x+7)^3
three because you subtracted one from the power
then you have the inside functions
4(x^2+3x+7)^3+2x+3
since seven is a constant the derivative is 0 (you don't have to write it)
basically use the chain rule hope this helped

Answer 4

CHAIN RULE!!!! =D

'kay so,

you take the derivative of the outside function (bring down the four and subtract one from the exponent)
4(x^2 + 3x + 7)^3

then multiply by the derivative of the inside funtion:

2x + 3

so you get:

(4(x^2 + 3x + 7)^3)(2x+3)

Answer 5

Let y = f(x), u = x^2 + 3x + 7. Then y = u^4
By the chain rule, dy/dx = dy/du * du/dx
= 4 u ^ 3 * (2x + 3)
= 4 (x^2 +3x + 7) ^ 3 * (2x + 3).
I don't think you need to expand that (unless your teacher is a sadist, that is).