Question

If we have a function as \[y = x\left[ {\left( {\cos \dfrac{x}{2}} \right) + \sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right) + \sin x} \right] + \dfrac{1}{2}\sqrt x \] , then find \[\dfrac{{dy}}{{dx}} = ?\]
A). \[\left( {1 + x} \right)\cos x + \left( {1 - x} \right)\sin x - \dfrac{1}{{4x\sqrt x }}\]
B). \[\left( {1 - x} \right)\cos x + \left( {1 - x} \right)\sin x + \dfrac{1}{{4x\sqrt x }}\]
C). \[\left( {1 + x} \right)\cos x + \left( {1 + x} \right)\sin x - \dfrac{1}{{4x\sqrt x }}\]
D). None of these

Answer 1

Hint: Here in the given question we need to solve the differentiation of the given expression, which contains trigonometric terms as well as simple variables, we know the associated formulae for both the terms, and accordingly will solve for the solution.
Formulae Used:
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x \\
   \Rightarrow \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\
   \Rightarrow \left( {{{\cos }^2}\dfrac{x}{2}} \right) - \left( {{{\sin }^2}\dfrac{x}{2}} \right) = \cos \left( {2 \times \dfrac{x}{2}} \right) = \cos x \\
   \Rightarrow \dfrac{d}{{dx}}\left( {u \times v} \right) = u'v + uv'\,\left( {u' = \dfrac{{du}}{{dx}}} \right) \\
 \]

Complete step-by-step solution:
Here in the given question we have a expression which need to be differentiated with respect to “x”, and contain terms of trigonometry as well as simple variables are also there, on solving we get:
We have:
\[\Rightarrow y = x\left[ {\left\{ {\left( {\cos \dfrac{x}{2}} \right) + \left( {\sin \dfrac{x}{2}} \right)} \right\}\left\{ {\left( {\cos \dfrac{x}{2}} \right) - \left( {\sin \dfrac{x}{2}} \right)} \right\} + \sin x} \right] + \dfrac{1}{2}\sqrt x \\
   \Rightarrow y = x\left[ {\left\{ {\left( {{{\cos }^2}\dfrac{x}{2}} \right) - \left( {{{\sin }^2}\dfrac{x}{2}} \right)} \right\} + \sin x} \right] + \dfrac{1}{2}\sqrt x \\
   \Rightarrow \dfrac{d}{{dx}}\left( {u \times v} \right) = u'v + uv'\,\left( {u' = \dfrac{{du}}{{dx}}} \right) \]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(x) \times \left[ {\left\{ {\left( {{{\cos }^2}\dfrac{x}{2}} \right) - \left( {{{\sin }^2}\dfrac{x}{2}} \right)} \right\} + \sin x} \right] + x\dfrac{d}{{dx}}\left[ {\left\{ {\left( {{{\cos }^2}\dfrac{x}{2}} \right) - \left( {{{\sin }^2}\dfrac{x}{2}} \right)} \right\} + \sin x} \right] + \dfrac{d}{{dx}}\left( {\dfrac{1}{2}\sqrt x } \right) \\
   \Rightarrow we\,know\,\left[ {\left( {{{\cos }^2}\dfrac{x}{2}} \right) - \left( {{{\sin }^2}\dfrac{x}{2}} \right) = \cos \left( {2 \times \dfrac{x}{2}} \right) = \cos x} \right] \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 1 \times \left[ {\cos x + \sin x} \right] + x\dfrac{d}{{dx}}\left[ {\cos x + \sin x} \right] + \dfrac{1}{{2 \times 2}}\left( {\dfrac{1}{{\sqrt x }}} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \left[ {\cos x + \sin x} \right] + x\left[ { - \sin x + \cos x} \right] + \dfrac{1}{{4\sqrt x }} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \cos x + \sin x - x\sin x + x\cos + \dfrac{1}{{4\sqrt x }} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = (1 + x)\cos x + (1 - x)\sin x + \dfrac{1}{{4\sqrt x }} \\
 \]
Here we get the required differential for the given expression, and this is our simplified answer for the given expression.

Additional Information: In the given question we first convert the given expansion in term of \[\cos 2x\] and then solve further, by using the pro

Note: Here in the given question, we first need to simplify the expression, in order to solve it fast. As here the given expression has expansion of trigonometric identity, hence need to replace it with required terms in order to solve further.