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Please check my deriving

  • Thread starter madeeeeee
  • Start date
  • #1
87
0
f(x)=(x^(2)+5x+2)^(4)

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
 

Answers and Replies

  • #2
statdad
Homework Helper
1,495
35
f(x)=(x^(2)+5x+2)^(4)

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

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

than my equation of the tangent turned out to be y=160x +16[/QUOTE]
 
  • #3
87
0
Im sorry, i don't understand where i went wrong
 
  • #4
87
0
can i leave it as f'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)
 
  • #5
eumyang
Homework Helper
1,347
10
f'(x)= 4(x^2 + 5x + 2)^3 (2x + 5)
= 8x + 20(x^2 + 5x + 2)^3
is this right
Notice something missing in your expression? Do you really want to multiply only the 20 by (x2 + 5x + 2)3?

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

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