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Differentiate cos^3(3x) ?
I got to the first stage but I'm stuck on the second and third stage:
1st stage: cos^3(3x) is same as (cos3x)^3 so let U=cos3x and dy^3/dx=u^3 differentiating dy^3/dx=u^3 would become 3u^2. Now put back the u in 3u^2...so 3(cos3x)^2 it is same as 3cos^2(3x)...
therefore 1st stage: 3cos^2(3x)
2nd stage : ?
3rd stage: ? (answer)
5 Answers
- 9 years ago
Can be re-written as:
(cos(3x))^3
Step 1: power rule
3(cos(3x)^2
Second step: differentiate cos(3x): = -sin(3x)
Third step differentiate 3x: = 3
las step: multiply them all together because you are using the chain rule:
=3((cos(3x))^2)*-sin(3x)*3
=-9((cos(3x))^2)*sin(3x)
- Anonymous9 years ago
Let u=3x so y=cos^3u, then let v=cosu and b=v^3 so v'=-sinu and b'=3v^2, times b' and v' to give -sinu3v^2 and sub in to give -sinu3cos^2u
Use this in first product differential to give u'=3 and y'=-3sinucos^2u and times them to give -9sinucos^2u
Finally sub back in to give -9sin(3x)cos^2(3x)
- 4 years ago
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