Question

# Find the second and third derivative of a function$y=\sin x$.

Hint: You can use the first principle method to find derivatives of $\sin x$ and
$\cos x$or directly use the differentiation of $\sin x$and $\cos x$wherever required.

We have the given function as –
$y=\sin x-(1)$
We know that differentiation of any function can be calculated with the help of first principle method of differentiation as stated below: -
If we have a function $f\left( x \right)$ which is continuous and differentiable for any real number then differentiation of it at any point $c$ can be stated as:
${{f}^{'}}(c)=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)-f(c)}{x-c}-(2)$
Applying LHL (left hand limit) and RHL (right hand limit) to equation (2) as,
For RHL:
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(c+h)-f(c)}{h}$
And LHL can be written as,
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(c-h)-f(c)}{-h}$
And we can verify that values got from LHL and RHL should be equal.
Hence, applying first principle method of differentiation we can find derivative of,
$y=\sin x$as $\cos x$.
Hence, $\dfrac{dy}{dx}=\dfrac{d}{dx}(\sin x)=\cos x-(3)$
Now, coming to the question part, we need to find the second and third derivative of the function. So, differentiating equation (3) again
$\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}(\cos x)=-\sin x$
(Differentiation of $\cos x$ can also be proved by first principle method of differentiation).
Hence, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\sin x-(4)$
Now, for the third derivative of the given function; differentiate equation (4) again, as: -
\begin{align} & \dfrac{d}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( -\sin x \right) \\ & \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( -\sin x \right) \\ \end{align}
Using equation (3) as $\dfrac{d}{dx}\left( \sin x \right)=\cos x$we can get,
$\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\cos x-(5)$
Therefore, derivative of $\sin x$ is $\cos x$and second and third derivative of $\sin x$ are
($-\sin x$) and ($-\cos x$) respectively from equation (4) and (5).
Note: One can go wrong while putting the values of $\dfrac{d}{dx}(\cos x)$ and $\dfrac{d}{dx}(\sin x)$. Student can write $\dfrac{d}{dx}(\sin x)=-\cos x$or $\dfrac{d}{dx}(\cos x)=\sin x$(confusion).
We can prove $\dfrac{d}{dx}(\sin x)=\cos x$by first principle method as written in the
solution. Let us prove it.
We have formula for first principle method as: -
${{f}^{'}}=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)-f(c)}{x-c}$
$f(x)=\sin x$and $f(c)=\sin c$
$\underset{x\to c}{\mathop{\lim }}\,\dfrac{\sin x-\sin c}{x-c}-(1)$
We have formula for $\sin C-\sin D$as
$\sin C-\sin D=2\sin \dfrac{C-D}{2}\cos \dfrac{C+D}{2}$
Hence,
$\sin x-\sin c=2\sin \dfrac{x-c}{2}\cos \dfrac{x+c}{2}$
Putting above value in equation (1)
\begin{align} & {{f}^{'}}(c)=\underset{x\to c}{\mathop{\lim }}\,\dfrac{2\sin \left( \dfrac{x-c}{2} \right)\cos \left( \dfrac{x+c}{2} \right)}{x-c} \\ & f'(c)=\underset{x\to c}{\mathop{\lim }}\,\dfrac{2\sin \dfrac{\left( x-c \right)}{2}}{2\left( \dfrac{x-c}{2} \right)}\cos \left( \dfrac{x+c}{2} \right) \\ & {{f}^{'}}(c)=\underset{x\to c}{\mathop{\lim }}\,\dfrac{\sin \dfrac{\left( x-c \right)}{2}}{\left( \dfrac{x-c}{2} \right)}\underset{x\to c}{\mathop{\lim }}\,\cos \left( \dfrac{x+c}{2} \right) \\ & {{f}^{'}}(c)=1\times \cos C=\cos C\left( \because \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 \right) \\ \end{align}
Hence proved