Question

Asked Feb 26, 2019

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find g'

g(x)= square root of 1-sin^2(x)

Step 1

The given expression is: g(x)= square root of 1-sin^2(x)

I am interpreting it to be as shown on the white board.

Step 2

Recall the famous formula of trigonometry: sin^{2}x + cos^{2}x = 1

Hence, 1 - sin^{2}x = cos^{2}x

Hence, the given expression takes the shape as shown on white board.

Step 3

Hence, g(x) = cosx

Recall that d(cosx) / dx = ...

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