Question

If $y=({\cos }^{-1}x{)}^2,$ then $(1-x^2)y_2-x{y}_1-2$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

First finding the differentiation of ${\cos }^{-1}\left(x\right)$ using first principle

Let $y^{\prime }$ = ${\cos }^{-1}\left(x\right)$

$x^{\prime }$ = $\cos \left(y^{\prime }\right)$

$\frac{df\left(y^{\prime }\right)}{d{y}^{\prime }}$ = ${lim}_{h\to 0}\frac{f(y^{\prime }+h)-f\left(y^{\prime }\right)}{h}$

$\frac{dx}{d{y}^{\prime }}$ = ${lim}_{h\to 0}\frac{\cos (y^{\prime }+h)-\cos \left(y^{\prime }\right)}{h}$

$\frac{dx}{d{y}^{\prime }}$ = ${lim}_{h\to 0}\frac{2\sin \left(\frac{y^{\prime }+y^{\prime }+h}{2}\right)\sin \left(\frac{y^{\prime }-y^{\prime }-h}{2}\right)}{h}$

Since, $\cos \left(A\right)-\cos \left(B\right)=2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)$

$\frac{dx}{d{y}^{\prime }}$ = ${lim}_{h\to 0}\frac{2\sin (y^{\prime }+\frac{h}{2})\sin \left(\frac{-h}{2}\right)}{h}$

By using $li{m}_{h\to 0}\frac{\sin \left(h\right)}{h}=1$, we get

$\frac{dx}{d{y}^{\prime }}$ = $-\sin \left(y^{\prime }\right)$

Using $sin\left(a\right)=\sqrt{1-{\cos }^2\left(a\right)}$

And $x$ = $\cos \left(y^{\prime }\right)$

we get:

$\frac{dx}{d{y}^{\prime }}$ = $-\sqrt{1-x^2}$

So we have $\frac{d{y}^{\prime }}{dx}$ = $\frac{-1}{\sqrt{1-x^2}}$

So Differentiation of ${\cos }^{-1}\left(x\right)$ is known to us.

Now using the formula of Differentiation:

If $y=\left(f\right(x){)}^n$

Than $\frac{dy}{dx}=n{f}^{\prime }\left(x\right)\left(f\right(x){)}^{n-1}$

we will also use $\frac{d\left(\frac{f\left(x\right)}{g\left(x\right)}\right)}{x}=\frac{g\left(x\right)f^{\prime }\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{\left(g\right(x){)}^2}$

Now we are ready to differentiate $y=\left({\cos }^{-1}\right(x){)}^2$

$y^{\prime }=\frac{-2{\cos }^{-1}\left(x\right)}{\sqrt{1-x^2}}$

Now Differentiating $y^{\prime }$

$y"=\frac{2+y^{\prime }x}{1-x^2}$

after rearrangement we will get $y"(1-x^2)-y^{\prime }x-2=0$

which is our final asnwer