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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

\[\cos x\]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

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