3-5-5
X
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CC
Recall in this problem that the derivative of cotangent X is
negative cosecant squared X.
And recall that the derivative of the square root of X is 1 / 2
square roots of X.
So when we look at this problem, we get our DYDX to be the
derivative cotangent X is negative cosecant squared X -,
7.
The derivative of the square root of X is 1 / 2 square roots
of X, or in this case 7 * 1 / 2 square roots of X.
And then the derivative of a constant is just zero.