 cuquerob21

2023-03-03

How to find the derivative of $y=x\left(\mathrm{arcsin}\right)\left({x}^{2}\right)$? inkljvn

How do you find the derivative of $y=x\mathrm{sin}{x}^{2}$?
$y=xa\mathrm{sin}{x}^{2}=\underset{u}{\underset{⏟}{u}}\cdot \underset{v}{\underset{⏟}{a\mathrm{sin}{x}^{2}}}⇒y=u\cdot v⇒{y}^{\prime }={u}^{\prime }v+u{v}^{\prime }$
$v=a\mathrm{sin}{x}^{2}⇒\underset{f\left[g\left(x\right)\right]}{\underset{⏟}{a\mathrm{sin}\stackrel{g\left(x\right)}{\stackrel{⏞}{{x}^{2}}}}}=f\left[g\left(x\right)\right]⇒{v}^{\prime }={f}^{\prime }\left[g\left(x\right)\right]\cdot {g}^{\prime }\left(x\right)←\text{chain rule}$
$f\left[g\left(x\right)\right]=a\mathrm{sin}{x}^{2}⇒{f}^{\prime }\left[g\left(x\right)\right]=\frac{1}{\sqrt{1-\left({x}^{2}{\right)}^{2}}}⇒{f}^{\prime }\left[g\left(x\right)\right]=\frac{1}{\sqrt{1-{x}^{4}}}$
$g\left(x\right)={x}^{2}⇒{g}^{\prime }\left(x\right)=2x$
Substituting:
${v}^{\prime }={f}^{\prime }\left[g\left(x\right)\right]\cdot {g}^{\prime }\left(x\right)=\frac{1}{\sqrt{1-{x}^{4}}}\cdot 2x⇒{v}^{\prime }=\frac{2x}{\sqrt{1-{x}^{4}}}$
Substituting u,v,u' and v' in y':
${y}^{\prime }={u}^{\prime }v+u{v}^{\prime }=\left(1\right)\left(a\mathrm{sin}{x}^{2}\right)+\left(x\right)\left(\frac{2x}{\sqrt{1-{x}^{4}}}\right)⇒{y}^{\prime }=a\mathrm{sin}{x}^{2}+\frac{2{x}^{2}}{\sqrt{1-{x}^{4}}}$
Answer: $y=xa\mathrm{sin}{x}^{2}⇒{y}^{\prime }=a\mathrm{sin}{x}^{2}+\frac{2{x}^{2}}{\sqrt{1-{x}^{4}}}$

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