The square root throws me off...
http://25.media.tumblr.com/tumblr_mb8p96…
http://25.media.tumblr.com/tumblr_mb8p96…
-
Chain rule, you can rewrite that problem as:
[1+cos^2(x)]^1/2
Can you take the derivative from here?
1/2[1+cos^2(x)]^-1/2 (-sin(x)cos(x)) = -(Sin(x)Cos(x))/(2[1+cos^2(x)]^1/2)
[1+cos^2(x)]^1/2
Can you take the derivative from here?
1/2[1+cos^2(x)]^-1/2 (-sin(x)cos(x)) = -(Sin(x)Cos(x))/(2[1+cos^2(x)]^1/2)
-
The square root is the same as being to the power of 1/2.
So we can rewrite that as:
(1 + cos^2(z))^(1/2)
And that becomes:
(1/2)(1 + cos^2(z))^(-1/2)
So we can rewrite that as:
(1 + cos^2(z))^(1/2)
And that becomes:
(1/2)(1 + cos^2(z))^(-1/2)