How to differentiate #y= ln^3(cos^2 √(1-x)#?

Question

1145 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-11T15:00:31

#(3ln^{2}(cos^{2}(sqrt(1-x)))tan(sqrt(1-x)))/(sqrt(1-x))#

You have to use the Chain Rule five times:

If #y=f(x)=ln^{3}(cos^{2}(sqrt(1-x)))#, then

#f'(x)=3ln^{2}(cos^{2}(sqrt(1-x))) * d/dx(ln(cos^{2}(sqrt{1-x})))#

#=3ln^[2}(cos^{2}(sqrt(1-x))) * 1/(cos^{2}(sqrt(1-x))) * d/dx(cos^{2}(sqrt(1-x)))#

#=(3ln^[2}(cos^{2}(1-x)))/(cos^{2}(sqrt(1-x))) * 2cos(sqrt(1-x)) * d/dx(cos(sqrt(1-x)))#

#=(6ln^[2}(cos^{2}(1-x)))/(cos(sqrt(1-x))) * (-sin(sqrt(1-x))) * d/dx((1-x)^[1/2})#

#=-6ln^[2}(cos^[2}(1-x)) tan(sqrt(1-x)) * 1/2 (1-x)^{-1/2} * d/dx(1-x)#

#=(3ln^{2}(cos^{2}(sqrt(1-x)))tan(sqrt(1-x)))/(sqrt(1-x))#