Answer
Verified
456.6k+ views
Hint: In this we can find the first few derivatives and then generalise these to get the nth derivative.
The given function is \[f(x)=\sin x\]in \[\left( 0,\dfrac{\pi }{2} \right)\]
In order to find the nth derivative, first, we shall consider 1st, 2nd,3rd and 4thderivatives of the given function.
Now we will find the first derivative of the given expression.
So, let us consider\[{{f}^{'}}(x)\].
\[{{f}^{'}}(x)=\dfrac{d}{dx}\left( \sin x \right)\]
We know derivative of $\sin x$ is $\cos x$ , so the above equation becomes,
\[{{f}^{'}}(x)=\cos x\]
Now we will write $f'(x)$ in terms of \[\sin x\].
We know, $\cos x=\sin (90+x)$, so the first derivative can be written as,
\[{{f}^{'}}(x)=\sin \left( 90+x \right)...........(i)\]
Now we will find the second derivative of the given expression.
So, let us consider\[{{f}^{'}}(x)\]and differentiating it will respect to $'x'$, we get
\[{{f}^{''}}(x)=\dfrac{d}{dx}\left( \cos x \right)\]
We know derivative of $\cos x$ is $-\sin x$ , so the above equation becomes,
\[{{f}^{''}}(x)=-\sin x\]
We know, $\sin (180+x)=-\sin x$, so the second derivative can be written as,
\[{{f}^{''}}(x)=\sin \left( 180+x \right)\]
Now we find the general term derivative we will try to write this in terms of first derivative, i.e.,
\[{{f}^{''}}(x)=\sin \left( 2\times 90+x \right)...........(ii)\]
Now we will find the third derivative of the given expression.
So, let us consider\[f''(x)\]and differentiating it will respect to $'x'$, we get
\[{{f}^{'''}}(x)=\dfrac{d}{dx}\left( -\sin x \right)\]
We know derivative of $\sin x$ is $\cos x$ , so the above equation becomes,
\[{{f}^{'''}}(x)=-\cos x\]
We know, $\sin (270+x)=-\cos x$, so the third derivative can be written as,
\[{{f}^{'''}}(x)=\sin \left( 270+x \right)\]
Now we find the general term derivative we will try to write $'270'$ in terms of $'90'$, we get
\[{{f}^{''}}(x)=\sin \left( 3\times 90+x \right)..........(iii)\]
Similarly, we will find the fourth derivative, we get
So, let us consider\[f'''(x)\]and differentiating it will respect to $'x'$, we get
\[{{f}^{IV}}(x)=\dfrac{d}{dx}\left( -\cos x \right)\]
We know derivative of $\cos x$ is $-\sin x$ , so the above equation becomes,
\[{{f}^{IV}}(x)=-(-\sin x)=\sin x\]
We know, $\sin (360+x)=\sin x$, so the fourth derivative can be written as,
\[{{f}^{IV}}(x)=\sin \left( 360+x \right)\]
Now we find the general term derivative we will try to write $'360'$ in terms of $'90'$, we get
\[{{f}^{IV}}(x)=\sin \left( 4\times 90+x \right)..........(iv)\]
Now we will write all the four derivative equations as below:
\[{{f}^{'}}(x)=\sin \left( 1\times 90+x \right)\]
\[{{f}^{''}}(x)=\sin \left( 2\times 90+x \right)\]
\[{{f}^{'''}}(x)=\sin \left( 3\times 90+x \right)\]
\[{{f}^{IV}}(x)=\sin \left( 4\times 90+x \right)\]
By observing the above four equations we can write the nth derivative as,
\[{{f}^{n}}(x)=\sin \left( n\times 90+x \right)\]
This is the required solution.
Note: In this problem, the main key is expressing all the derivatives into \[\sin x\] function in the given range \[\left( 0,\dfrac{\pi }{2} \right)\]. In this way we can observe the similarity between the derivatives in order to find nth derivative.
The given function is \[f(x)=\sin x\]in \[\left( 0,\dfrac{\pi }{2} \right)\]
In order to find the nth derivative, first, we shall consider 1st, 2nd,3rd and 4thderivatives of the given function.
Now we will find the first derivative of the given expression.
So, let us consider\[{{f}^{'}}(x)\].
\[{{f}^{'}}(x)=\dfrac{d}{dx}\left( \sin x \right)\]
We know derivative of $\sin x$ is $\cos x$ , so the above equation becomes,
\[{{f}^{'}}(x)=\cos x\]
Now we will write $f'(x)$ in terms of \[\sin x\].
We know, $\cos x=\sin (90+x)$, so the first derivative can be written as,
\[{{f}^{'}}(x)=\sin \left( 90+x \right)...........(i)\]
Now we will find the second derivative of the given expression.
So, let us consider\[{{f}^{'}}(x)\]and differentiating it will respect to $'x'$, we get
\[{{f}^{''}}(x)=\dfrac{d}{dx}\left( \cos x \right)\]
We know derivative of $\cos x$ is $-\sin x$ , so the above equation becomes,
\[{{f}^{''}}(x)=-\sin x\]
We know, $\sin (180+x)=-\sin x$, so the second derivative can be written as,
\[{{f}^{''}}(x)=\sin \left( 180+x \right)\]
Now we find the general term derivative we will try to write this in terms of first derivative, i.e.,
\[{{f}^{''}}(x)=\sin \left( 2\times 90+x \right)...........(ii)\]
Now we will find the third derivative of the given expression.
So, let us consider\[f''(x)\]and differentiating it will respect to $'x'$, we get
\[{{f}^{'''}}(x)=\dfrac{d}{dx}\left( -\sin x \right)\]
We know derivative of $\sin x$ is $\cos x$ , so the above equation becomes,
\[{{f}^{'''}}(x)=-\cos x\]
We know, $\sin (270+x)=-\cos x$, so the third derivative can be written as,
\[{{f}^{'''}}(x)=\sin \left( 270+x \right)\]
Now we find the general term derivative we will try to write $'270'$ in terms of $'90'$, we get
\[{{f}^{''}}(x)=\sin \left( 3\times 90+x \right)..........(iii)\]
Similarly, we will find the fourth derivative, we get
So, let us consider\[f'''(x)\]and differentiating it will respect to $'x'$, we get
\[{{f}^{IV}}(x)=\dfrac{d}{dx}\left( -\cos x \right)\]
We know derivative of $\cos x$ is $-\sin x$ , so the above equation becomes,
\[{{f}^{IV}}(x)=-(-\sin x)=\sin x\]
We know, $\sin (360+x)=\sin x$, so the fourth derivative can be written as,
\[{{f}^{IV}}(x)=\sin \left( 360+x \right)\]
Now we find the general term derivative we will try to write $'360'$ in terms of $'90'$, we get
\[{{f}^{IV}}(x)=\sin \left( 4\times 90+x \right)..........(iv)\]
Now we will write all the four derivative equations as below:
\[{{f}^{'}}(x)=\sin \left( 1\times 90+x \right)\]
\[{{f}^{''}}(x)=\sin \left( 2\times 90+x \right)\]
\[{{f}^{'''}}(x)=\sin \left( 3\times 90+x \right)\]
\[{{f}^{IV}}(x)=\sin \left( 4\times 90+x \right)\]
By observing the above four equations we can write the nth derivative as,
\[{{f}^{n}}(x)=\sin \left( n\times 90+x \right)\]
This is the required solution.
Note: In this problem, the main key is expressing all the derivatives into \[\sin x\] function in the given range \[\left( 0,\dfrac{\pi }{2} \right)\]. In this way we can observe the similarity between the derivatives in order to find nth derivative.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE