найдите f'1 если f x=(2x^3+5x^2-7)^17

yes I can see now :)

f'(x)=17*(6x^2+10x).(2x^3+5x^2-7).^16

omg I forget again:p

I did it thnx for your helpin' :)

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