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

Question

2058 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-09T08:31:40

Use of chain rule twice and at the second derivative use of quotent rule.

First derivative

#2sin(lnx)*cos(lnx)*1/x#

Second derivative

#(2cos(2lnx)-sin(2lnx))/x^2#

First derivative

#(sin^2(lnx))'#

#2sin(lnx)*(sin(lnx))'#

#2sin(lnx)*cos(lnx)(lnx)'#

#2sin(lnx)*cos(lnx)*1/x#

Although this is acceptable, to make the second derivative easier, one can use the trigonometric identity:

#2sinθcosθ=sin(2θ)#

Therefore:

#(sin^2(lnx))'=sin(2lnx)/x#

Second derivative

#(sin(2lnx)/x)'#

#(sin(2lnx)'x-sin(2lnx)(x)')/x^2#

#(cos(2lnx)(2lnx)'x-sin(2lnx)*1)/x^2#

#(cos(2lnx)*2*1/x*x-sin(2lnx))/x^2#

#(2cos(2lnx)-sin(2lnx))/x^2#