Question

The $\dfrac{{d\left( {{{\cos }^{ - 1}}\sqrt {\cos x} } \right)}}{{dx}} =$
A. $\dfrac{1}{2}\sqrt {\sec x + 1}$
B. $\sqrt {\sec x + 1}$
C. $- \dfrac{1}{2}\sqrt {\sec x + 1}$
D. $- \sqrt {\sec x + 1}$

Answer 1

We are given with a function ${\cos ^{ - 1}}\sqrt {\cos x}$, we need to differentiate it with respect to $x$, to get the results. And for that we would use the chain rule of differentiation, in which we would go step by step differentiating the values and then rejoin all the values in order to get the desired result of the function.

Formula used:
$\dfrac{{d\left( {{{\cos }^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$
$\Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
$\Rightarrow \dfrac{{d\cos x}}{{dx}} = - \sin x$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dv}} \times \dfrac{{dv}}{{du}} \times \dfrac{{du}}{{dx}}$
$\Rightarrow \dfrac{1}{{\cos x}} = \sec x$
$\Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
$\Rightarrow {1^2} - {\cos ^2}x = \left( {1 - \cos x} \right)\left( {1 + \cos x} \right)$

Complete step by step answer:
We are given with a function ${\cos ^{ - 1}}\sqrt {\cos x}$, considering the value of ${\cos ^{ - 1}}\sqrt {\cos x}$ to be $y$, so the function becomes:
$y = {\cos ^{ - 1}}\sqrt {\cos x}$ ….(1)
Now, considering $\cos x$ to be u, and $y$ becomes:
$y = {\cos ^{ - 1}}\sqrt u$
Considering $\sqrt u$ to be v, and $y$ becomes:
$y = {\cos ^{ - 1}}v$
Now, differentiating $y$ with respect to v, we get:
$\dfrac{{dy}}{{dv}} = \dfrac{{d\left( {{{\cos }^{ - 1}}v} \right)}}{{dv}}$
From the formula of differentiation, we know that $\dfrac{{d\left( {{{\cos }^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$, so substituting this in the above value, we get:
$\dfrac{{dy}}{{dv}} = \dfrac{{ - 1}}{{\sqrt {1 - {v^2}} }}$
Substituting the value of v, we get:
$\dfrac{{dy}}{{dv}} = \dfrac{{ - 1}}{{\sqrt {1 - {{\left( {\sqrt u } \right)}^2}} }} = \dfrac{{ - 1}}{{\sqrt {1 - u} }}$
Substituting the value of u, we get:
$\dfrac{{dy}}{{dv}} = \dfrac{{ - 1}}{{\sqrt {1 - u} }} = \dfrac{{ - 1}}{{\sqrt {1 - \cos x} }}$ ……..(2)

Since, we had $v = \sqrt u$, so differentiating $v$ with respect to $u$, we get:
$\dfrac{{dv}}{{du}} = \dfrac{{d\sqrt u }}{{du}} = \dfrac{{d{u^{\dfrac{1}{2}}}}}{{du}}$
Since, we know that $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$, according to this, our value becomes:
$\dfrac{{dv}}{{du}} = \dfrac{1}{2}{u^{\dfrac{1}{2} - 1}} = \dfrac{1}{2}{u^{ - \dfrac{1}{2}}} = \dfrac{1}{{2{u^{\dfrac{1}{2}}}}} = \dfrac{1}{{2\sqrt u }}$
$\Rightarrow \dfrac{{dv}}{{du}} = \dfrac{1}{{2\sqrt u }}$
Substituting the value of u, we get:
$\Rightarrow \dfrac{{dv}}{{du}} = \dfrac{1}{{2\sqrt {\cos x} }}$ ……..(3)
Now, we also had $u = \cos x$, so differentiating $u$ with respect to $x$, we get:
$\dfrac{{du}}{{dx}} = \dfrac{{d\cos x}}{{dx}}$
Since, we know that $\dfrac{{d\cos x}}{{dx}} = - \sin x$, according to this, our value becomes:
$\dfrac{{du}}{{dx}} = - \sin x$ ……..(4)
Now, for the equation 1, $y = {\cos ^{ - 1}}\sqrt {\cos x}$, the chain rule of differentiation can be written as:
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dv}} \times \dfrac{{dv}}{{du}} \times \dfrac{{du}}{{dx}}$
Since, we have all the differentiated values, so substituting equation 2,3 and 4 in the above equation and we get:
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dv}} \times \dfrac{{dv}}{{du}} \times \dfrac{{du}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{\sqrt {1 - \cos x} }} \times \dfrac{1}{{2\sqrt {\cos x} }} \times \left( { - \sin x} \right) \\\$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin x}}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }}$ …..(5)

From the trigonometric identities, we know that ${\sin ^2}x + {\cos ^2}x = 1$:
So, subtracting both sides by ${\cos ^2}x$:
${\sin ^2}x + {\cos ^2}x - {\cos ^2}x = 1 - {\cos ^2}x$
$\Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Taking square root both the sides:
$\Rightarrow \sqrt {{{\sin }^2}x} = \sqrt {1 - {{\cos }^2}x} \\\ \Rightarrow \sin x = \sqrt {1 - {{\cos }^2}x} \\\$
Substituting the value of $\sin x$ in equation 5:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {{\cos }^2}x} }}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {{1^2} - {{\cos }^2}x} }}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }}$
Since, we can see that ${1^2} - {\cos ^2}x$ is in the form of ${a^2} - {b^2}$ that can be splitted as ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, similarly for ${1^2} - {\cos ^2}x = \left( {1 - \cos x} \right)\left( {1 + \cos x} \right)$.

Now, the equation becomes:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)} }}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }}$
Splitting out the roots:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {\left( {1 - \cos x} \right)} \sqrt {\left( {1 + \cos x} \right)} }}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }}$
Cancelling the common terms:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {\left( {1 + \cos x} \right)} }}{{2\sqrt {\cos x} }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{{\sqrt {\left( {1 + \cos x} \right)} }}{{\sqrt {\cos x} }}$
Taking roots common:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\sqrt {\dfrac{{1 + \cos x}}{{\cos x}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\sqrt {\dfrac{1}{{\cos x}} + \dfrac{{\cos x}}{{\cos x}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\sqrt {\dfrac{1}{{\cos x}} + 1}$
Since, we know that $\dfrac{1}{{\cos x}} = \sec x$, so the value becomes:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\sqrt {\sec x + 1}$
Therefore, the result is $\dfrac{{d{{\cos }^{ - 1}}\sqrt {\cos x} }}{{dx}} = \dfrac{1}{2}\sqrt {\sec x + 1}$.

Hence, option A is correct.

Note: The chain rule used for this question is $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dv}} \times \dfrac{{dv}}{{du}} \times \dfrac{{du}}{{dx}}$, similarly for other functions, we would shorter or longer chains in the same method. Remember to solve the differentiation step by step in order to avoid mistakes and for correct solving. It’s important to remember and implement the correct formulas in the places needed, a slight difference in the formula may lead to huge changes in the function.