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I am unable to get this sum. Find $\frac{dy}{dx}$. This is the Question as a picture. Here is the Mathjax of the equation: $y = \arcsin(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2})$

In the answer key that I have, it has been directly simplified to: $[\arcsin(x) - \arcsin(rt x) ]$ (I am unable to understand how this was obtained). and then it has been differentiated.

Jose M Serra
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user329952
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  • Use https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y and https://math.stackexchange.com/questions/2099342/why-does-arcsin-x-arcsinx – lab bhattacharjee Jan 03 '19 at 14:48

2 Answers2

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HINT: Use the chain rule $$\frac{d}{dx} f(g(x))=g'(x)f'(g(x))$$ with $f(x)=\arcsin(x)$ and $g(x)=x\sqrt{1-x}-\sqrt{x(1-x^2)}$. You can use the product rule to differentiate $g(x)$.

aleden
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To answer your question thoroughly I would insist you learn the chain rule, and the product rule to answer your question. I will highlight what the chain rule is and what the product rule is, and when they are being used, orange is both. \begin{array}{|c|c|}\hline \bbox[yellow]{\text{Chain Rule}}&\frac{d}{dx}f(u(x))= \frac{d}{du}f(u(x))\frac{du}{dx}u(x)=f'(u(x))u'(x) \\ \hline \bbox[red]{\text{Product Rule}}& \frac{d}{dx}(u(x)v(x))=u(x)v'(x)+u'(x)v(x)\\ \hline \end{array} For simplicity I will start the problem highlighting the initial problem as being a usage of chain rule. \begin{align}&\frac{d}{dx}(\bbox[yellow]{\arcsin(x\sqrt{1-x}-\sqrt{x-x^2})}\\ &\frac{1}{1-(x\sqrt{1-x^2}-\sqrt{x-x^2})^2}(\frac{d}{dx}(\bbox[orange]{x\sqrt{1-x}}-\bbox[yellow]{\sqrt{x-x^2}})\\ &\frac{d}{dx}(\bbox[yellow]{\arcsin(x\sqrt{1-x}-\sqrt{x-x^2})}=\frac{\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}-\frac{1-3x^2}{2\sqrt{x-x^3}}}{\sqrt{1-\left(x\sqrt{\left(1-x\right)}-\sqrt{x-x^3}\right)^2}}\end{align}

Jose M Serra
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