How do you find the derivative of #sin^2(lnx)#?

1 Answer
Jim H
Nov 1, 2015

Recall that #sin^2(lnx) = [sin(lnx)]^2# and use the chain rule twice.

Explanation:

#d/dx(sin^2(lnx)) =d/dx( [sin(lnx)]^2)#

# = 2sin(lnx) [d/dx(sin(lnx))]#

# = 2sin(lnx) [cos(lnx)d/dx(lnx)]#

# = 2sin(lnx) cos(lnx)[1/x]#

# = (2sin(lnx) cos(lnx))/x#

Which we may prefer to write as:

# = sin(2lnx)/x#

Or, perhaps as

# = sin(ln(x^2))/x#

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