Question

If $y=\sin ^{-1}\left(\dfrac{1-x^{2}}{1+x^{2}}\right)$ , then $\mathrm{dy} / \mathrm{dx}$ equals
(1) $2 /\left(1-x^{2}\right)$
(2) $1 /\left(1+x^{2}\right)$
(3) $-2 /\left(1+x^{2}\right)$
(4) $2 /\left(1-x^{2}\right)$

Answer 1

Hint: Here we are differentiating the given expression to find the required value. The rate at which one variable changes with respect to another is known as differentiation.

Formula Used:
If $y = f(x)$ and $f(x)$ is differentiable, then $f'(x)$ or $dy/dx$ is used to indicate the differentiation.

Complete step by step Solution:
Given that
$y=\sin ^{-1}\left(\dfrac{1-x^{2}}{1+x^{2}}\right)$
differentiating
$=\dfrac{1}{\sqrt{1-\left(\dfrac{1-x^{2}}{1+x^{2}}\right)^{2}}} \dfrac{d}{d x}\left(\dfrac{1-x^{2}}{1+x^{2}}\right)$
$\dfrac{d}{d x}\left(\dfrac{1-x^{2}}{1+x^{2}}\right)=-\dfrac{4 x}{\left(1+x^{2}\right)^{2}}$
Then substitute the value
$=\dfrac{1}{\sqrt{1-\left(\dfrac{1-x^{2}}{1+x^{2}}\right)^{2}}}\left(-\dfrac{4 x}{\left(1+x^{2}\right)^{2}}\right)$
$=\dfrac{1+x^{2}}{2 x}\left(-\dfrac{4 x}{\left(1+x^{2}\right)^{2}}\right)$
Apply rule: $\quad a(-b)=-a b$
$\dfrac{1+x^{2}}{2 x}\left(-\dfrac{4 x}{\left(1+x^{2}\right)^{2}}\right)=-\dfrac{1+x^{2}}{2 x} \cdot \dfrac{4 x}{\left(1+x^{2}\right)^{2}}$
$=-\dfrac{1+x^{2}}{2 x} \cdot \dfrac{4 x}{\left(1+x^{2}\right)^{2}}$
Apply the fraction rule: $\dfrac{a}{b} \cdot \dfrac{c}{d}=\dfrac{a \cdot c}{b \cdot d}$
$=-\dfrac{\left(1+x^{2}\right) \cdot 4 x}{2 x\left(1+x^{2}\right)^{2}}$
Cancel the common factor: $x$
$=-\dfrac{\left(1+x^{2}\right)-4}{2\left(1+x^{2}\right)^{2}}$
Factor the number: $4=2 \cdot 2$
$=-\dfrac{\left(1+x^{2}\right) \cdot 2 \cdot 2}{2\left(1+x^{2}\right)^{2}}$
Cancel the common factor: 2
$=-\dfrac{\left(1+x^{2}\right) \cdot 2}{\left(1+x^{2}\right)^{2}}$
The way of representing huge numbers in terms of powers is known as an exponent. The amount of times an integer has been repeated by itself is the exponent. For instance, the number 6 is multiplied by itself four times, yielding $6 \times 6 \times 6$. You can write this as $6^{4}$.
 In this case, the exponent is $b+c$and the base is $\quad a^{b+c}=a^{b} \cdot a^{c}$ .
Apply exponent rule: $\quad a^{b+c}=a^{b} \cdot a^{c}$
$\left(1+x^{2}\right)^{2}=\left(1+x^{2}\right)\left(1+x^{2}\right)$
$=-\dfrac{\left(1+x^{2}\right) \cdot 2}{\left(1+x^{2}\right)\left(1+x^{2}\right)}$
Cancel the common factor: $1+x^{2}$
$=-\dfrac{2}{1+x^{2}}$

Hence, the correct option is 3.

Note: Differentiation is the ratio of a slight change in one quantity to a little change in another that depends on the first quantity. Calculus' main emphasis on the differentiation of a function makes it one of the subject's key ideas. Differentiation is the process of determining the maximum or lowest value of a function, the speed and acceleration of moving objects, and the tangent of a curve. If $y = f(x)$ and $f(x)$ is differentiable, then $f'(x)$ or $dy/dx$ is used to indicate the differentiation.