Question

Find $\dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right)$.
A) $4\cos ec\left( {2x} \right) \times \cot \left( {2x} \right)$
B) $ - 4\cos ec\left( {2x} \right) \times \cot \left( {2x} \right)$
C) $ - 4\cos ec\left( x \right) \times \cot \left( {2x} \right)$
D) None of these

Answer 1

Hint: We are required to differentiate a trigonometric function involving cosecant and secant functions. So, we first convert the secant and cosecant functions to sine and cosine to simplify the expression. Then, we use the half angle formula for sine to convert the function into easily integrable form and integrate the function by substitution.

Complete step by step solution: 
Let us assume this function as $f\left( x \right)$.
So, we have, $f\left( x \right) = \sqrt {{{\sec }^2}x + \cos e{c^2}x} $.
Now, we use the trigonometric formulae for cosecant and secant as $\sec x = \dfrac{1}{{\cos x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$. Hence, substituting $\sec x$ and $\cos ecx$, we get,
$ \Rightarrow f\left( x \right) = \sqrt {\dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}}} $
Taking LCM in the square root, we get,
$ \Rightarrow f\left( x \right) = \sqrt {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} $
Using the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, we get,
$ \Rightarrow f\left( x \right) = \sqrt {\dfrac{1}{{{{\sin }^2}x{{\cos }^2}x}}} $
\[ \Rightarrow f\left( x \right) = \dfrac{1}{{\sin x\cos x}}\]
Multiplying numerator and denominator by 2, we get,
\[ \Rightarrow f\left( x \right) = \dfrac{2}{{2\sin x\cos x}}\]
Using double angle formula for sine $\sin 2x = 2\sin x\cos x$,
\[ \Rightarrow f\left( x \right) = \dfrac{2}{{\sin 2x}}\]
\[ \Rightarrow f\left( x \right) = 2\cos ec\left( {2x} \right)\]
Now, $\dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = \dfrac{d}{{dx}}\left( {2\cos ec\left( {2x} \right)} \right)$
Taking constant out of differentiation. We get,
$\dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = 2\dfrac{d}{{dx}}\left( {\cos ec\left( {2x} \right)} \right)$
Let $t = 2x$. Using the chain rule of differentiation, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = 2\dfrac{d}{{dt}}\left( {\cos ec\left( t \right)} \right) \times \dfrac{{dt}}{{dx}}$
We know the differentiation of \[\cos ecx\] is \[ - \cos ecx \times \cot x\].
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = 2\left[ { - \cos ec\left( t \right) \times \cot \left( t \right)} \right] \times \dfrac{{dt}}{{dx}}$
Substituting t in the expression,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = 2\left[ { - \cos ec\left( {2x} \right) \times \cot \left( {2x} \right)} \right] \times \dfrac{{d\left( {2x} \right)}}{{dx}}$
Using the power rule of differentiation, we get $\dfrac{{d\left( {2x} \right)}}{{dx}} = 2$,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = 2\left[ { - \cos ec\left( {2x} \right) \times \cot \left( {2x} \right)} \right] \times 2$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {{{\sec }^2}x + \cos e{c^2}x} } \right) = - 4\cos ec\left( {2x} \right) \times \cot \left( {2x} \right)$
So, the correct answer is option (B) $ - 4\cos ec\left( {2x} \right) \times \cot \left( {2x} \right)$.

Note: The trigonometric formulae such as $\sec x = \dfrac{1}{{\cos x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$ should be remembered to solve such questions. We should remember to substitute the variable introduced back at the final stage in order to get the correct answer. Chain rule of differentiation is followed for differentiating composite functions.