Question

How do you differentiate $y = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ ?

Answer 1

Hint: In this question, we are given a trigonometric equation and we have been asked to differentiate the question. There are two methods to solve this question. Both are as follows:
Method 1: We can directly differentiate the given equation using inverse trigonometric formulae and chain rule.
Method 2: in this method, we will shift the trigonometric ratio to the other side and we will put down the inverse trigonometry. Now, differentiate this equation. After differentiating, convert every in terms of x. Do not leave the answer in terms of y.

Complete step by step answer:
We will solve using method 1.
$y = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ …. (given)
While differentiating this equation, we will use chain rule. It is used in cases when there is a function inside the function. In this case, our outer function is ${\sin ^{ - 1}}x$ and the inner function is $\dfrac{1}{x}$ . We will name the outer function as $f\left( x \right)$ an inner function as $g\left( x \right)$ . We apply chain rule as follows:
$ \dfrac{{d\left( {f\left( {g\left( x \right)} \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$
Using the chain rule, now we will differentiate both the sides with respect to x,
It will give us,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{1}{x}} \right)}^2}} }} \times \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$ ….. (As $\dfrac{{d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$)
We can also write $\dfrac{1}{x}$ as ${x^{ - 1}}$ . Simplifying the denominator,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\dfrac{{{x^2} - 1}}{{{x^2}}}} }} \times \dfrac{{d\left( {{x^{ - 1}}} \right)}}{{dx}}$
Simplifying further and using $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ ,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\dfrac{{\sqrt {{x^2} - 1} }}{x}}} \times \left( { - {x^{ - 1 - 1}}} \right)$
Flipping the denominator and simplifying the second term,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{{\sqrt {{x^2} - 1} }} \times \dfrac{{ - 1}}{{{x^2}}}$
Eliminating the like terms,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{x\sqrt {{x^2} - 1} }}$
Hence, differentiation of $y = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ is $\dfrac{{ - 1}}{{x\sqrt {{x^2} - 1} }}$ .

Note: Now, we will use method 2 to find the answer.
In this method, we will shift the trigonometric ratio to the other side and close the inverse trigonometry.
$ \Rightarrow y = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ …. (given)
Shifting sin to the other side,
$\sin y = \left( {\dfrac{1}{x}} \right)$ ….. (1)
Now, we will differentiate the equation with respect to x,
\[ \Rightarrow \cos y\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^2}}}\] …... $\left( {\dfrac{{d\left( {\sin y} \right)}}{{dx}} = \cos y} \right)$
Shifting the ratio to the other side,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^2}\cos y}}\] …. (2)
We cannot leave our answer in terms of y as our question was in terms of x. So, we will use certain trigonometric formulae to convert the answer.
We know that ${\sin ^2}y + {\cos ^2}y = 1$ .
Using it to find the value of $\cos y$ ,
$ \Rightarrow \cos y = \sqrt {1 - {{\sin }^2}y} $
From equation (1), we know that $\sin y = \left( {\dfrac{1}{x}} \right)$ .
Substituting in the above equation,
$ \Rightarrow \cos y = \sqrt {1 - \dfrac{1}{{{x^2}}}} $
Putting it back in equation (2),
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^2}\sqrt {1 - \dfrac{1}{{{x^2}}}} }}\]
On simplifying, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^2}\dfrac{{\sqrt {{x^2} - 1} }}{x}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{x\sqrt {{x^2} - 1} }}\]
Hence, the answer using this formula is similar to the answer using the above formula.