How do you differentiate y = ((sin(x))^6 (tan(x))^2) / (x^2 + 2)^2?

Apr 13, 2017

By Log Differentiation,

$y ' = \left(8 \cot x + 2 \tan x - \frac{4 x}{{x}^{2} + 2}\right) \frac{{\sin}^{6} x \cdot {\tan}^{2} x}{{x}^{2} + 2} ^ 2$

Explanation:

You can also use Logarithmic Differentiation.

By $\tan x = \sin \frac{x}{\cos} x$,

y=(sin^6x cdot tan^2 x)/(x^2+2)^2 =(sin^6x cdot (sin^2 x)/(cos^2 x))/(x^2+2)^2 =(sin^8 x)/(cos^2x(x^2+2)^2)

By taking the natural log of both sides,

$R i g h t a r r o w \ln y = \ln \left(\frac{{\sin}^{8} x}{{\cos}^{2} x {\left({x}^{2} + 2\right)}^{2}}\right)$

By Log Properties: $\ln \left(x \cdot y\right) = \ln x + \ln y$ and $\ln \left(\frac{x}{y}\right) = \ln x - \ln y ,$

$R i g h t a r r o w \ln y = \ln \left({\sin}^{8} x\right) - \ln \left({\cos}^{2} x\right) - \ln {\left({x}^{2} + 2\right)}^{2}$

By Log Property: $\ln {x}^{r} = r \ln x$,

$R i g h t a r r o w \ln y = 8 \ln \left(\sin x\right) - 2 \ln \left(\cos x\right) - 2 \ln \left({x}^{2} + 2\right)$

By differentiating with respect to $x$ using $\left[\ln \left(g \left(x\right)\right)\right] ' = \frac{g ' \left(x\right)}{g} \left(x\right)$,

$R i g h t a r r o w \frac{y '}{y} = 8 \cos \frac{x}{\sin} x - 2 \frac{- \sin x}{\cos x} - 2 \frac{2 x}{{x}^{2} + 2}$

By cleaning up a bit,

$R i g h t a r r o w \frac{y '}{y} = 8 \cot x + 2 \tan x - \frac{4 x}{{x}^{2} + 2}$

By multiplying both sides by $y$,

$R i g h t a r r o w y ' = \left(8 \cot x + 2 \tan x - \frac{4 x}{{x}^{2} + 2}\right) \frac{{\sin}^{6} x \cdot {\tan}^{2} x}{{x}^{2} + 2} ^ 2$

I hope that this was clear.