If $f\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$, then $f'\left( {\dfrac{\pi }{4}} \right)$ is equal to
A) $\sqrt 2 $
B) $\dfrac{1}{{\sqrt 2 }}$
C) $0$
D) $\dfrac{{\sqrt 3 }}{2}$
Answer
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Hint: In order to find the value of $f'\left( {\dfrac{\pi }{4}} \right)$, we need to solve the function $f\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$ to its simplest term possible by dividing the function by 2sinx and continue it, then differentiate the function with respect to x. Substitute the value of $x = \dfrac{\pi }{4}$, simplify and get the results.
Formula used:
$2\sin x\cos x = \sin 2x$
$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$
$\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$\cos 8\pi = 1{\text{ and sin8}}\pi = 0$
Complete step by step answer:
We are given with the function, $f\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$ and we need to solve this function in order to find the value of $f'\left( {\dfrac{\pi }{4}} \right)$.
Let’s start with the function $f\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$.
Multiplying and dividing the function by $2\sin x$, we get:
$f\left( x \right) = \dfrac{{2\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2\sin x}}$
From the trigonometric angles and sub- angle’s, we know that $2\sin x\cos x = \sin 2x$. So, substituting this value in the above function, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2\sin x}}$
Again, multiplying and dividing the value by 2, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2 \times 2\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{4\sin x}}$ ……(1)
We can see there is $2\sin 2x\cos 2x$, which can be written as:
$2\sin 2x\cos 2x = \sin 2\left( {2x} \right) = \sin 4x$
Substituting the above value in the equation 1 and we get:
$f\left( x \right) = \dfrac{{\sin 4x\cos 4x\cos 8x\cos 16x}}{{4\sin x}}$
Multiplying and dividing the value by 2, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 4x\cos 4x\cos 8x\cos 16x}}{{2 \times 4\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 4x\cos 4x\cos 8x\cos 16x}}{{8\sin x}}$ ……(2)
We can see again there is $2\sin 4x\cos 4x$, which can be written as:
$2\sin 4x\cos 4x = \sin 2\left( {4x} \right) = \sin 8x$
Substituting the above value in the equation 2 and we get:
$f\left( x \right) = \dfrac{{\sin 8x\cos 8x\cos 16x}}{{8\sin x}}$
These steps to be followed again.
So, again multiplying and dividing the function by 2, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 8x\cos 8x\cos 16x}}{{2 \times 8\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 8x\cos 8x\cos 16x}}{{16\sin x}}$
Again, we get $2\sin 8x\cos 8x$, which can be written as $2\sin 8x\cos 8x = \sin 2\left( {8x} \right) = \sin 16x$.
Substituting this value in the above function, we get:
$f\left( x \right) = \dfrac{{\sin 16x\cos 16x}}{{16\sin x}}$
One last time multiplying and dividing the function by 2:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 16x\cos 16x}}{{2 \times 16\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 16x\cos 16x}}{{32\sin x}}$
Writing $2\sin 16x\cos 16x = \sin 2\left( {16x} \right) = \sin 32x$ in the above function, we get:
$f\left( x \right) = \dfrac{{\sin 32x}}{{32\sin x}}$ …….(3)
And, we can see that it cannot be further solved.
Now, since we wanted $f'\left( x \right)$, so differentiating equation 3 with respect to x and we get:
$f'\left( x \right) = \dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}}$ ……(4)
Since, the function is in the form of quotient rule of differentiation, that is $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}}$ which when differentiated, we get:
$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$ …………(5)
Comparing equation 4 with $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$, we get:
$\dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}} = \dfrac{{32\sin x\left( {\dfrac{{d\left( {\sin 32x} \right)}}{{dx}}} \right) - \sin 32x\left( {\dfrac{{d\left( {32\sin x} \right)}}{{dx}}} \right)}}{{{{\left( {32\sin x} \right)}^2}}}$ .……..(6)
Now,
Comparing equation 4 with $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}}$, we get:
$u = \sin 32x$
$v = 32\sin x$
So, differentiating them we get:
$\dfrac{{du}}{{dx}} = \dfrac{{d\left( {\sin 32x} \right)}}{{dx}} = 32\cos 32x$
$\dfrac{{dv}}{{dx}} = \dfrac{{d\left( {32\sin x} \right)}}{{dx}} = 32\cos x$
Substituting these values in equation 6, we get:
$ \Rightarrow \dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}} = \dfrac{{32\sin x\left( {32\cos 32x} \right) - \sin 32x\left( {32\cos x} \right)}}{{{{\left( {32\sin x} \right)}^2}}}$
Simplifying it we get:
$ \Rightarrow \dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}} = \dfrac{{{{\left( {32} \right)}^2}\sin x\cos 32x - 32\cos x\sin 32x}}{{{{\left( {32\sin x} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{{{\left( {32} \right)}^2}\sin x\cos 32x - 32\cos x\sin 32x}}{{{{\left( {32\sin x} \right)}^2}}}$
Substituting $x = \dfrac{\pi }{4}$:
$ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{{\left( {32} \right)}^2}\sin \dfrac{\pi }{4}\cos 32\left( {\dfrac{\pi }{4}} \right) - 32\cos \dfrac{\pi }{4}\sin 32\left( {\dfrac{\pi }{4}} \right)}}{{{{\left( {32\sin \dfrac{\pi }{4}} \right)}^2}}}$
$ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{{\left( {32} \right)}^2}\sin \dfrac{\pi }{4}\cos 8\pi - 32\cos \dfrac{\pi }{4}\sin 8\pi }}{{{{\left( {32\sin \dfrac{\pi }{4}} \right)}^2}}}$
Since, we know that $\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos 8\pi = 1{\text{ and sin8}}\pi = 0$.
So, we get:
$ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{\left( {{{\left( {32} \right)}^2} \times \dfrac{1}{{\sqrt 2 }} \times 1} \right) - \left( {32 \times \dfrac{1}{{\sqrt 2 }} \times 0} \right)}}{{{{\left( {32 \times \dfrac{1}{{\sqrt 2 }}} \right)}^2}}}$
\[ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{{\left( {32} \right)}^2} \times \dfrac{1}{{\sqrt 2 }}}}{{{{\left( {32} \right)}^2} \times {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}}}\]
Cancelling the similar terms, we get:
\[ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}} = \sqrt 2 \]
\[f'\left( {\dfrac{\pi }{4}} \right) = \sqrt 2 \]
Hence, Option (A) is correct.
Note:
1. Remember to solve the questions step by step, because we can see that every step followed was the same, so to avoid confusion or any error use this method.
2. It’s important to know the correct formula that will be used, the wrong formula at the wrong step can give wrong results.
Formula used:
$2\sin x\cos x = \sin 2x$
$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$
$\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$\cos 8\pi = 1{\text{ and sin8}}\pi = 0$
Complete step by step answer:
We are given with the function, $f\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$ and we need to solve this function in order to find the value of $f'\left( {\dfrac{\pi }{4}} \right)$.
Let’s start with the function $f\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$.
Multiplying and dividing the function by $2\sin x$, we get:
$f\left( x \right) = \dfrac{{2\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2\sin x}}$
From the trigonometric angles and sub- angle’s, we know that $2\sin x\cos x = \sin 2x$. So, substituting this value in the above function, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2\sin x}}$
Again, multiplying and dividing the value by 2, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2 \times 2\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{4\sin x}}$ ……(1)
We can see there is $2\sin 2x\cos 2x$, which can be written as:
$2\sin 2x\cos 2x = \sin 2\left( {2x} \right) = \sin 4x$
Substituting the above value in the equation 1 and we get:
$f\left( x \right) = \dfrac{{\sin 4x\cos 4x\cos 8x\cos 16x}}{{4\sin x}}$
Multiplying and dividing the value by 2, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 4x\cos 4x\cos 8x\cos 16x}}{{2 \times 4\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 4x\cos 4x\cos 8x\cos 16x}}{{8\sin x}}$ ……(2)
We can see again there is $2\sin 4x\cos 4x$, which can be written as:
$2\sin 4x\cos 4x = \sin 2\left( {4x} \right) = \sin 8x$
Substituting the above value in the equation 2 and we get:
$f\left( x \right) = \dfrac{{\sin 8x\cos 8x\cos 16x}}{{8\sin x}}$
These steps to be followed again.
So, again multiplying and dividing the function by 2, we get:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 8x\cos 8x\cos 16x}}{{2 \times 8\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 8x\cos 8x\cos 16x}}{{16\sin x}}$
Again, we get $2\sin 8x\cos 8x$, which can be written as $2\sin 8x\cos 8x = \sin 2\left( {8x} \right) = \sin 16x$.
Substituting this value in the above function, we get:
$f\left( x \right) = \dfrac{{\sin 16x\cos 16x}}{{16\sin x}}$
One last time multiplying and dividing the function by 2:
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 16x\cos 16x}}{{2 \times 16\sin x}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2\sin 16x\cos 16x}}{{32\sin x}}$
Writing $2\sin 16x\cos 16x = \sin 2\left( {16x} \right) = \sin 32x$ in the above function, we get:
$f\left( x \right) = \dfrac{{\sin 32x}}{{32\sin x}}$ …….(3)
And, we can see that it cannot be further solved.
Now, since we wanted $f'\left( x \right)$, so differentiating equation 3 with respect to x and we get:
$f'\left( x \right) = \dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}}$ ……(4)
Since, the function is in the form of quotient rule of differentiation, that is $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}}$ which when differentiated, we get:
$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$ …………(5)
Comparing equation 4 with $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$, we get:
$\dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}} = \dfrac{{32\sin x\left( {\dfrac{{d\left( {\sin 32x} \right)}}{{dx}}} \right) - \sin 32x\left( {\dfrac{{d\left( {32\sin x} \right)}}{{dx}}} \right)}}{{{{\left( {32\sin x} \right)}^2}}}$ .……..(6)
Now,
Comparing equation 4 with $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}}$, we get:
$u = \sin 32x$
$v = 32\sin x$
So, differentiating them we get:
$\dfrac{{du}}{{dx}} = \dfrac{{d\left( {\sin 32x} \right)}}{{dx}} = 32\cos 32x$
$\dfrac{{dv}}{{dx}} = \dfrac{{d\left( {32\sin x} \right)}}{{dx}} = 32\cos x$
Substituting these values in equation 6, we get:
$ \Rightarrow \dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}} = \dfrac{{32\sin x\left( {32\cos 32x} \right) - \sin 32x\left( {32\cos x} \right)}}{{{{\left( {32\sin x} \right)}^2}}}$
Simplifying it we get:
$ \Rightarrow \dfrac{{d\left( {\dfrac{{\sin 32x}}{{32\sin x}}} \right)}}{{dx}} = \dfrac{{{{\left( {32} \right)}^2}\sin x\cos 32x - 32\cos x\sin 32x}}{{{{\left( {32\sin x} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{{{\left( {32} \right)}^2}\sin x\cos 32x - 32\cos x\sin 32x}}{{{{\left( {32\sin x} \right)}^2}}}$
Substituting $x = \dfrac{\pi }{4}$:
$ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{{\left( {32} \right)}^2}\sin \dfrac{\pi }{4}\cos 32\left( {\dfrac{\pi }{4}} \right) - 32\cos \dfrac{\pi }{4}\sin 32\left( {\dfrac{\pi }{4}} \right)}}{{{{\left( {32\sin \dfrac{\pi }{4}} \right)}^2}}}$
$ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{{\left( {32} \right)}^2}\sin \dfrac{\pi }{4}\cos 8\pi - 32\cos \dfrac{\pi }{4}\sin 8\pi }}{{{{\left( {32\sin \dfrac{\pi }{4}} \right)}^2}}}$
Since, we know that $\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos 8\pi = 1{\text{ and sin8}}\pi = 0$.
So, we get:
$ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{\left( {{{\left( {32} \right)}^2} \times \dfrac{1}{{\sqrt 2 }} \times 1} \right) - \left( {32 \times \dfrac{1}{{\sqrt 2 }} \times 0} \right)}}{{{{\left( {32 \times \dfrac{1}{{\sqrt 2 }}} \right)}^2}}}$
\[ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{{\left( {32} \right)}^2} \times \dfrac{1}{{\sqrt 2 }}}}{{{{\left( {32} \right)}^2} \times {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}}}\]
Cancelling the similar terms, we get:
\[ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}} = \sqrt 2 \]
\[f'\left( {\dfrac{\pi }{4}} \right) = \sqrt 2 \]
Hence, Option (A) is correct.
Note:
1. Remember to solve the questions step by step, because we can see that every step followed was the same, so to avoid confusion or any error use this method.
2. It’s important to know the correct formula that will be used, the wrong formula at the wrong step can give wrong results.
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