Find the second and third derivative of a function\[y=\sin x\].
Answer
365.1k+ views
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
Last updated date: 01st Oct 2023
•
Total views: 365.1k
•
Views today: 9.65k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

How many millions make a billion class 6 maths CBSE

How many crores make 10 million class 7 maths CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
