Answer
Verified
426.9k+ 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
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Assertion CNG is a better fuel than petrol Reason It class 11 chemistry CBSE
How does pressure exerted by solid and a fluid differ class 8 physics CBSE
Number of valence electrons in Chlorine ion are a 16 class 11 chemistry CBSE
What are agricultural practices? Define
What does CNG stand for and why is it considered to class 10 chemistry CBSE
The rate of evaporation depends on a Surface area b class 9 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
State whether the following statement is true or false class 11 physics CBSE
A night bird owl can see very well in the night but class 12 physics CBSE