- #1

madeeeeee

- 87

- 0

f'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)

= 8x + 20(x^2 + 5x + 2)^3

is this right

than my equation of the tangent turned out to be y=160x +16

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- Thread starter madeeeeee
- Start date

- #1

madeeeeee

- 87

- 0

f'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)

= 8x + 20(x^2 + 5x + 2)^3

is this right

than my equation of the tangent turned out to be y=160x +16

- #2

statdad

Homework Helper

- 1,495

- 36

f(x)=(x^(2)+5x+2)^(4)

f'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)

Ok

Not okay.= 8x + 20(x^2 + 5x + 2)^3

is this right

than my equation of the tangent turned out to be y=160x +16[/QUOTE]

- #3

madeeeeee

- 87

- 0

Im sorry, i don't understand where i went wrong

- #4

madeeeeee

- 87

- 0

can i leave it as f'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)

- #5

eumyang

Homework Helper

- 1,347

- 11

Notice something missing in your expression? Do you really want to multiply only the 20 by (xf'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)

= 8x + 20(x^2 + 5x + 2)^3

is this right

Where did you get this? Was there a point that was given?than my equation of the tangent turned out to be y=160x +16

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