How to differentiate $y = {ln}^{3} ({cos}^{2} \sqrt (1 - x)$ $y= ln^3(cos^2 √(1-x)$ ?

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Answer 1 · 2015-08-11T15:00:31

$\frac{3 {ln}^{2} ({cos}^{2} (\sqrt{1 - x})) tan (\sqrt{1 - x})}{\sqrt{1 - x}}$ $(3ln^{2}(cos^{2}(sqrt(1-x)))tan(sqrt(1-x)))/(sqrt(1-x))$

You have to use the Chain Rule five times:

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

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

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

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

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

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

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

How to differentiate y= ln^3(cos^2 √(1-x)y=ln3(cos2√(1−x)y= ln^3(cos^2 √(1-x)?