Question

How do you find the fourth derivative of \[\sin x\]?

Answer 1

Hint: From the question given, we have been asked to find the fourth derivative of \[\sin x\].We can find the fourth derivative of \[\sin x\] by using the differentiation formula. First we have to find the first derivative and then the second derivative and then the third derivative and at last we have to find the fourth derivative.

Complete step by step answer:
From the question given, we have to find the fourth derivative of \[\sin x\].
As we have already discussed above, first we have to find the first derivative.
First derivative of \[\sin x\] is
Let us assume that \[y=\sin x\]
To obtain the first derivative, we have to differentiate \[y\].
Then first derivative will be
\[\Rightarrow {{y}_{1}}=\dfrac{d}{dx}\left( \sin x \right)\]
\[\Rightarrow {{y}_{1}}=\cos x\]
Now, we have to find the second derivative. To find the second derivative, we have to differentiate the first derivative which we got above.
So, the second derivative will be,
\[\Rightarrow {{y}_{2}}=\dfrac{d}{dx}\left( \cos x \right)\]
\[\Rightarrow {{y}_{2}}=-\sin x\]
Now, we have to find the third derivative. To find the third derivative, we have to differentiate the second derivative which we got above.
So, the third derivative will be,
\[\Rightarrow {{y}_{3}}=\dfrac{d}{dx}\left( -\sin x \right)\]
\[\Rightarrow {{y}_{3}}=-\cos x\]
Now, we have to find the fourth derivative. To find the fourth derivative, we have to differentiate the third derivative which we got above.
So, the fourth derivative will be,
\[\Rightarrow {{y}_{4}}=\dfrac{d}{dx}\left( -\cos x \right)\]
\[\Rightarrow {{y}_{4}}=\sin x\]
Therefore, the fourth derivative is \[{{y}_{4}}=\sin x\].

Note: We should be very careful while finding the first derivative because it is the base step to find the fourth derivative for the given question. Also, we should be very careful while finding the remaining derivatives. Also, we should be well aware of the differentiation and finding derivatives. Similarly we can find the fourth derivative of $\cos x$ as follows
 $\begin{align}
  & {{y}_{1}}=\dfrac{d}{dx}\cos x\Rightarrow -\sin x \\
 & \Rightarrow {{y}_{2}}=\dfrac{d}{dx}\left( -\sin x \right)\Rightarrow -\cos x \\
 & \Rightarrow {{y}_{3}}=\dfrac{d}{dx}\left( -\cos x \right)\Rightarrow \sin x \\
 & \Rightarrow {{y}_{4}}=\dfrac{d}{dx}\sin x\Rightarrow \cos x \\
\end{align}$