Question

Find $\dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right]$.

Answer 1

Hint: In the given problem, we are required to differentiate a function involving trigonometric and inverse trigonometric functions. We first simplify the expression of the given function. Then, we use the rules of differentiation in order to differentiate the function. The power rule of differentiation must be known in order to solve the question.

Formula Used: $2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1 = \cos x$
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step answer:
Let us assume this function as $f\left( x \right)$.
So, we have, $f\left( x \right) = {\cos ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)$.
Now, we know the half angle formula for cosine as $2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1 = \cos x$ . Hence, substituting $\cos x$ as $2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right] = \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + 2{{\cos }^2}\left( {\dfrac{x}{2}} \right) - 1}}{2}} } \right)} \right]$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right] = \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}{2}} } \right)} \right]$
Simplifying the expression, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right] = \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {{{\cos }^2}\left( {\dfrac{x}{2}} \right)} } \right)} \right]$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right] = \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\cos \left( {\dfrac{x}{2}} \right)} \right)} \right]$
Now, we know that ${\cos ^{ - 1}}\left( {\cos x} \right) = x$. So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right] = \dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)$
Using power rule of differentiation $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)} \right] = \dfrac{1}{2}$

Note: The trigonometric formulae such as the half-angle formula for cosine should be remembered to solve such questions. We should go through the differentiation rules such as the power rule, product rule and quotient rule to do the differentiation of functions. The chain rule of differentiation is followed for differentiating composite functions.