Question

Sin inverse x under root 1 minus x minus root x under root 1 - x square

Answer 1

Given that,

$y=\sin^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})$

Suppose, $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x(1-x)}}+p$

Let x= sin θ[/tex]

Then, $\cos\theta=\sqrt{1-x^2}$

Let,$\sqrt{x}=\sin\phi$

Then, $\cos\phi=\sqrt{1-x}$

We need to calculate the value of p

Using given equation

$y=\sin^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})$

Put the value in the equation

$y=\sin^{-1}(\sin\theta\cos\phi+\sin\phi\cos\theta)$

$y=\sin^{-1}\sin(\theta+\phi)$

$y=\theta+\phi$

$y=\sin^{-1}x+\sin^{-1}\sqrt{x}$

On differentiating

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{\sqrt{1-x}}\times\dfrac{1}{2\sqrt{x}}$

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{2\sqrt{x(1-x)}}$

Now comparing the equation with given equation

$\dfrac{dy}{dx}=p+\dfrac{1}{2\sqrt{x}(1-x)}$

So, the value of p will be

$p= \dfrac{1}{\sqrt{1-x^2}}$

Hence, The value of p is $\dfrac{1}{\sqrt{1-x^2}}$.