128k views
5 votes
Find dy/dx for y=log (tan2x)

User Aneer Geek
by
8.7k points

1 Answer

1 vote

Answer:


(dy)/(dx) = (2)/(sin2x cos2x)

Explanation:

y = log (tan 2x)

Applying chain rule:


(dy)/(dx) = (d)/(dx) (1st function) . (d)/(dx) (2nd function). (d)/(dx)(3rd function)


(dy)/(dx) = (d)/(dx) log(tan 2x) . (d)/(dx) tan2x. (d)/(dx) 2x


(dy)/(dx) = (1)/(tan2x) * sec^(2)2x * 2


(dy)/(dx) = (2sec^(2)2x )/(tan2x)

Now since
sec^(2)x = (1)/(cos^(2)x ) \ and\ tanx = (sinx)/(cosx)


(dy)/(dx) = (2cos2x )/(sin2x cos^22x)


(dy)/(dx) = (2)/(sin2x cos2x)

User Rishabh Poddar
by
8.3k points

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