Question

The first derivative of the function $\sin 2x\cos 2x\cos 3x + {\log _2}{2^{x + 3}}$ with respect to x at $x = \pi $ is:
(A) $2$
(B) $ - 1$
(C) $1$
(D) None of these

Answer 1

Hint: In the given problem, we are required to differentiate the function $\sin 2x\cos 2x\cos 3x + {\log _2}{2^{x + 3}}$ with respect to x and find the value of derivative for $x = \pi $. Since, $\sin 2x\cos 2x\cos 3x + {\log _2}{2^{x + 3}}$ is a complex function involving a product term, so we will produce a rule of differentiation. We will first simplify the trigonometric term with the help of the double angle formula of sine.

Complete step-by-step solution:
So, we have, $\sin 2x\cos 2x\cos 3x + {\log _2}{2^{x + 3}}$
Multiplying and dividing the first term by two, we get,
$ \Rightarrow \left( {\dfrac{{2\sin 2x\cos 2x}}{2}} \right)\cos 3x + {\log _2}{2^{x + 3}}$
Using the double angle formula for sine $\sin 2x = 2\sin x\cos x$, we get,
$ \Rightarrow \dfrac{1}{2}\sin 4x\cos 3x + {\log _2}{2^{x + 3}}$
Now, we use the law of logarithm ${\log _a}{x^n} = n{\log _a}x$. So, we get,
$ \Rightarrow \dfrac{1}{2}\sin 4x\cos 3x + \left( {x + 3} \right){\log _2}2$
We know that the value of ${\log _2}2$ is $1$. So, we get,
\[ \Rightarrow \dfrac{1}{2}\sin 4x\cos 3x + \left( {x + 3} \right)\]
So, now we have to differentiate the simplified function $\dfrac{1}{2}\sin 4x\cos 3x + x + 3$.
$\dfrac{d}{{dx}}\left( {\dfrac{1}{2}\sin 4x\cos 3x + x + 3} \right)$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{2}\sin 4x\cos 3x} \right) + \dfrac{d}{{dx}}\left[ x \right] + \dfrac{d}{{dx}}\left[ 3 \right]$
We know that the derivative of a constant term is zero. Also, we know the power rule of differentiation as $\dfrac{{d\left[ {{x^n}} \right]}}{{dx}} = n{x^{\left( {n - 1} \right)}}$. Taking the constant out of the differentiation, we get,
$ \Rightarrow \dfrac{1}{2}\dfrac{d}{{dx}}\left( {\sin 4x\cos 3x} \right) + 1 + 0$
Now, applying the product rule of differentiation \[\dfrac{d}{{dx}}\left[ {f\left( x \right) \times g\left( x \right)} \right] = f\left( x \right)\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}} + g\left( x \right)\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}\].
$ \Rightarrow \dfrac{1}{2}\left[ {\sin 4x\dfrac{d}{{dx}}\left( {\cos 3x} \right) + \left( {\cos 3x} \right)\dfrac{d}{{dx}}\left( {\sin 4x} \right)} \right] + 1$
We know that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $ - \sin x$. So, applying the chain rule of differentiation in the expression, we get,
$ \Rightarrow \dfrac{1}{2}\left[ {\sin 4x \times \left( { - 3\sin 3x} \right) + \left( {\cos 3x} \right)\left( {4\cos 4x} \right)} \right] + 1$
$ \Rightarrow \dfrac{1}{2}\left[ { - 3\sin 3x\sin 4x + 4\cos 4x\cos 3x} \right] + 1$
Opening the brackets and simplifying the expression, we get,
$ \Rightarrow \dfrac{{ - 3\sin 3x\sin 4x}}{2} + 2\cos 4x\cos 3x + 1$
Now, we have to put the value of x as $\pi $ in the first derivative of the function. So, we get,
$ \Rightarrow \dfrac{{ - 3\sin \left( {3\pi } \right)\sin \left( {4\pi } \right)}}{2} + 2\cos \left( {4\pi } \right)\cos \left( {3\pi } \right) + 1$
Putting the values of trigonometric functions, we get,
$ \Rightarrow \dfrac{{ - 3 \times 0 \times 0}}{2} + 2 \times 1 \times \left( { - 1} \right) + 1$
Simplifying the calculations, we get,
$ \Rightarrow 0 - 2 + 1$
$ \Rightarrow - 1$
So, option (B) is the correct answer.

Note: We must know the various rules of differentiation such as product rule, chain rule and quotient rule to solve such problems. We should take care while substituting the values of the variable and carrying out the calculations so as to be sure of the final answer. We should also know the properties of logarithms to get to the required solution.