Question

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

Answer 1

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: -

& \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{aligned}$$ Using equation (3) as $$\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{aligned} & {{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{aligned}$$ Hence proved