How do you differentiate $f (x) = {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠ \cdot cos 2 x$ $f(x)=sin^2(sqrt(3x/(x-1)^2))*cos2x$ using the chain rule?

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How do you differentiate $f (x) = {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠ \cdot cos 2 x$ $f(x)=sin^2(sqrt(3x/(x-1)^2))*cos2x$ using the chain rule?

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Answer 1 · 2017-03-30T14:44:33

$\frac{d f}{d x} = - \frac{\sqrt{3}}{2 \sqrt{x}} sin ⎛ ⎝ \sqrt{\frac{12 x}{{(x - 1)}^{2}}} ⎞ ⎠ \frac{(x + 1)}{{(x - 1)}^{2}} \cdot cos 2 x - {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠ sin 2 x$ $(df)/(dx)=-sqrt3/(2sqrtx)sin(sqrt((12x)/(x-1)^2))((x+1))/(x-1)^2*cos2x-sin^2(sqrt(3x/(x-1)^2))sin2x$

Explanation:

Let us first differentiate $g (x) = {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠$ $g(x)=sin^2(sqrt(3x/(x-1)^2))$

let us write this as chain, where $g (x) = {sin}^{2} (h (x))$ $g(x)=sin^2(h(x))$ , $h (x) = \sqrt{p (x)}$ $h(x)=sqrt(p(x))$ ,

$p (x) = \frac{3 x}{{(x - 1)}^{2}}$ $p(x)=(3x)/(x-1)^2$ and using chain rule

$\frac{d g (x)}{d x} = \frac{d g}{d h} \times \frac{d h}{d p} \times \frac{d p}{d x}$ $(dg(x))/(dx)=(dg)/(dh)xx(dh)/(dp)xx(dp)/(dx)$

= $2 sin (h (x)) cos (h (x)) \times \frac{1}{2 \sqrt{p (x)}} \times \frac{3 {(x - 1)}^{2} - 6 x (x - 1)}{{(x - 1)}^{4}}$ $2sin(h(x))cos(h(x))xx1/(2sqrt(p(x)))xx(3(x-1)^2-6x(x-1))/(x-1)^4$

= $sin (2 h (x)) \times \frac{1}{2 \sqrt{p (x)}} \times \frac{3 x^{2} - 6 x + 3 - 6 x^{2} + 6 x}{{(x - 1)}^{4}}$ $sin(2h(x))xx1/(2sqrt(p(x)))xx(3x^2-6x+3-6x^2+6x)/(x-1)^4$

= $sin ⎛ ⎝ \sqrt{\frac{12 x}{{(x - 1)}^{2}}} ⎞ ⎠ \times \frac{1}{2 \sqrt{\frac{3 x}{{(x - 1)}^{2}}}} \times \frac{- 3 x^{2} + 3}{{(x - 1)}^{4}}$ $sin(sqrt((12x)/(x-1)^2))xx1/(2sqrt((3x)/(x-1)^2))xx(-3x^2+3)/(x-1)^4$

= $- 3 sin ⎛ ⎝ \sqrt{\frac{12 x}{{(x - 1)}^{2}}} ⎞ ⎠ \times \frac{1}{2 \sqrt{3 x}} \times \frac{x^{2} - 1}{{(x - 1)}^{3}}$ $-3sin(sqrt((12x)/(x-1)^2))xx1/(2sqrt(3x))xx(x^2-1)/(x-1)^3$

= $- \frac{\sqrt{3}}{2 \sqrt{x}} sin ⎛ ⎝ \sqrt{\frac{12 x}{{(x - 1)}^{2}}} ⎞ ⎠ \frac{(x + 1)}{{(x - 1)}^{2}}$ $-sqrt3/(2sqrtx)sin(sqrt((12x)/(x-1)^2))((x+1))/(x-1)^2$

and hence

$\frac{d f}{d x} = - \frac{\sqrt{3}}{2 \sqrt{x}} sin ⎛ ⎝ \sqrt{\frac{12 x}{{(x - 1)}^{2}}} ⎞ ⎠ \frac{(x + 1)}{{(x - 1)}^{2}} \cdot cos 2 x - {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠ sin 2 x$ $(df)/(dx)=-sqrt3/(2sqrtx)sin(sqrt((12x)/(x-1)^2))((x+1))/(x-1)^2*cos2x-sin^2(sqrt(3x/(x-1)^2))sin2x$

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How do you differentiate $f (x) = {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠ \cdot cos 2 x$ $f(x)=sin^2(sqrt(3x/(x-1)^2))*cos2x$ using the chain rule?

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Explanation:

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How do you differentiate $f (x) = {sin}^{2} ⎛ ⎜ ⎝ \sqrt{3 \frac{x}{{(x - 1)}^{2}}} ⎞ ⎟ ⎠ \cdot cos 2 x$ $f(x)=sin^2(sqrt(3x/(x-1)^2))*cos2x$ using the chain rule?