Question

Assertion (A): The ${{n}^{\text{th}}}$ derivative of $\sin 5x\cos 3x$ is $\dfrac{1}{2}\left[ {{8}^{n}}\sin \left( \dfrac{n\pi }{2}+8x \right)+{{2}^{n}}\sin \left( \dfrac{n\pi }{2}+2x \right) \right]$
Reason (R): If $y=\sin \left( ax+b \right)$ then ${{y}_{n}}={{a}^{n}}\sin \left( ax+b+\dfrac{n\pi }{2} \right)$
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer 1

Hint: Here, in this question, we have to first we will use the Assertion statement, use the trigonometric transformation, $\sin \text{A}\cdot \text{cosB}\,\text{= }\dfrac{1}{2}\left[ \sin \left( \text{A + B} \right)+\sin \left( \text{A}-\text{B} \right) \right]$ to simplify the $y$ given in the Assertion statement, then let us use the formula mentioned in the Reason statement and check if the proof satisfies due to the reason statement.

Complete step by step answer:
Here, in this question, we have been given a proof to prove in the first statement that is the assertion(A), and we have also been given a second statement which is the reason (R). Let us find out if both these statements, Assertion and Reason are related to each other.
Firstly, we have been given $y=\sin 5x\cos 3x$, hence let us find its ${{n}^{\text{th}}}$ derivative.
We have, $y=\sin 5x\cos 3x$
Now, from the trigonometric transformations we know,
$\sin \text{A}\cdot \text{cosB}\,\text{= }\dfrac{1}{2}\left[ \sin \left( \text{A + B} \right)+\sin \left( \text{A}-\text{B} \right) \right]$
By using the above, formula and $y=\sin 5x\cos 3x$ we get,
$\begin{align}
  & y=\dfrac{1}{2}\left[ \sin \left( 5x+3x \right)+\sin \left( 5x-3x \right) \right] \\
 & =\dfrac{1}{2}\left[ \sin 8x+\sin 2x \right]
\end{align}$
We can write, $\sin 8x=\sin \left( 8x+0 \right)$ and $\sin 2x=\sin \left( 2x+0 \right)$. Also, let us find the ${{n}^{\text{th}}}$derivative by differentiating $y$ with respect to $x$.
We know, the ${{n}^{\text{th}}}$ derivative for $y=\sin \left( ax+b \right)$ is ${{y}_{n}}={{a}^{n}}\sin \left( ax+b+\dfrac{n\pi }{2} \right)$.
Now, after we differentiate $y$, we get
${{y}_{n}}=\dfrac{1}{2}\left[ \dfrac{{{d}^{n}}}{d{{x}^{n}}}\sin \left( 8x+0 \right)+\dfrac{{{d}^{n}}}{d{{x}^{n}}}\sin \left( 2x+0 \right) \right]$
   $=\dfrac{1}{2}\left[ {{8}^{n}}\sin \left( 8x+0+\dfrac{n\pi }{2} \right)+{{2}^{n}}\sin \left( 2x+0+\dfrac{n\pi }{2} \right) \right]$
After further simplifying the above expression, we will get
${{y}_{n}}=\dfrac{1}{2}\left[ {{8}^{n}}\sin \left( 8x+\dfrac{n\pi }{2} \right)+{{2}^{n}}\sin \left( 2x+\dfrac{n\pi }{2} \right) \right]$
We got the above proof, that the ${{n}^{\text{th}}}$ derivative of $\sin 5x\cos 3x$ is ${{y}_{n}}=\dfrac{1}{2}\left[ {{8}^{n}}\sin \left( 8x+\dfrac{n\pi }{2} \right)+{{2}^{n}}\sin \left( 2x+\dfrac{n\pi }{2} \right) \right]$
But the proof is satisfied with the help of the ${{n}^{\text{th}}}$ derivative for $y=\sin \left( ax+b \right)$ is ${{y}_{n}}={{a}^{n}}\sin \left( ax+b+\dfrac{n\pi }{2} \right)$.

So, the correct answer is “Option A”.

Note: This question is in the form of successive differentiation. Successive differentiation is the process of differentiating the given function successively ‘n’ times which will give us successive derivatives. The ${{n}^{\text{th}}}$ derivative is shown by $\dfrac{{{d}^{n}}x}{d{{x}^{n}}}$, for n = 1, 2, 3, and so on, these can also be denoted as ${{f}^{n}}\left( x \right)$. For example, for second order derivative, you can either use $\dfrac{{{d}^{2}}x}{d{{x}^{2}}}$ or ${{f}’’}\left( x \right)$.