Найдите f'1 если f x=(2x^3+5x^2-7)^17
I hope this helps you f'(x)=17.(6x^2+10x).(2x^3+5x^2-7).^16 x=1 f'(1)=17.(6.1^2+10.1)(2.1^3+5.1^2-7)^16 f'(1)=17.16.0 f'(1)=0
oh you can delete my answer :) sorry
your final answer is correct
:) thnx you're nice
Correct your answer.
Under the answer there is a button "fix" ( "исправить" )
yes I can see now :)
f'(x)=17*(6x^2+10x).(2x^3+5x^2-7).^16
omg I forget again:p
You can fix it again
I did it thnx for your helpin' :)