Deriving f(x)=(x^2+5x+2)^4 & Finding Tangent with y=160x+16

[madeeeeee]

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

 

[statdad]

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]

 

[madeeeeee]

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

 

[madeeeeee]

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

 

[eumyang]

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?