Question

I have the answer can someone please check me...
The derivative of f(x)=3(cos(2x+4))^7 is
f'(x)=ksin(2x+4)(cos(2x+4))^6 For what value of k?

Answer 1

\[f(x)=3(cos(2x+4))^7 \]
\[f'(x)=21\cos(2x+4)^6\sin(2x+4)\times 2=42\cos(2x+4)^6\sin(2x+4)\]

Answer 2

oops i am off by a minus sign

Answer 3

\[f'(x)=21\cos(2x+4)^6\times -\sin(2x+4)\times 2=-42\cos(2x+4)^6\sin(2x+4)\]

Answer 4

So then the value of k is 3

Answer 5

i think u r doing little wrong

Answer 6

no, \(k=-42\)

Answer 7

oh ok I understqand I put it into my calculator wrong I just redidi it and I go the right answer now that matches yours

Answer 8

it will b\[f \prime =-3 \times 7 \cos ^{6}(2x+4) \sin (2x+4) \times 2\]

Answer 9

did u c my solution @satellite73