Question 4a(ii)

SCH_CALC L3CALC

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$$y={\sin }^{-1}(x)⟺x=\sin y$$

Now differentiate implicitly:

$$1=\cos y\cdot \frac{dy}{dx}$$ $$\frac{dy}{dx}=\frac{1}{\cos y}$$ $$x=\sin y⟺x^2={\sin }^{2}y⟺x^2=1-{\cos }^{2}y⟺\cos y=\pm \sqrt{1-x^2}$$ $$\frac{dy}{dx}=\frac{1}{\pm \sqrt{1-x^2}}$$

However, ${\sin }^{-1}x$ is defined within $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$, and since $\cos y$ from $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$ is always positive, use the positive root!

Hence

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}∎$$

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For minimum, $g^{\prime }(x)=0$.

$$g^{\prime }(x)=\frac{d}{dx}({\sin }^{-1}2x)+\frac{d}{dx}\left({\sin }^{-1}\left(\frac{\pi }{4}-2x\right)\right)$$ $$g^{\prime }(x)=\frac{1}{\sqrt{1-(2x{)}^2}}\cdot (2)+\frac{1}{\sqrt{1-{\left(\frac{\pi }{4}-2x\right)}^2}}\cdot (-2)$$ $$g^{\prime }(x)=\frac{2}{\sqrt{1-(2x{)}^2}}-\frac{2}{\sqrt{1-{\left(\frac{\pi }{4}-2x\right)}^2}}=0=\frac{2}{0}$$ $$⟹\sqrt{1-(2x{)}^2}-\sqrt{1-{\left(\frac{\pi }{4}-2x\right)}^2}=0$$ $$\sqrt{1-(2x{)}^2}=\sqrt{1-{\left(\frac{\pi }{4}-2x\right)}^2}$$ $$1-(2x{)}^2=1-{\left(\frac{\pi }{4}-2x\right)}^2$$ $$(2x{)}^2={\left(\frac{\pi }{4}-2x\right)}^2$$ $$\pm 2x=\frac{\pi }{4}-2x$$ $$2x\pm 2x=\frac{\pi }{4}$$ $$4x=\frac{\pi }{4}or0=\frac{\pi }{4}$$ $$0\ne \frac{\pi }{4}⟹4x=\frac{\pi }{4}$$

Hence

$$\boxed{x=\frac{\pi }{16}}$$