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?

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?

anonymous · Answer

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

anonymous · Answer

oops i am off by a minus sign

anonymous · Answer

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

anonymous · Answer

So then the value of k is 3

shamim · Answer

i think u r doing little wrong

anonymous · Answer

no, $k=-42$

anonymous · Answer

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

shamim · Answer

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

shamim · Answer

did u c my solution @satellite73