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

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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

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

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

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

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

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

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