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Question: Find the nth derivative of the following expression: \[f(x)=\sin x\]in \[\left( 0,\dfrac{\pi }{2} ...

Find the nth derivative of the following expression:
f(x)=sin⁡xf(x)=\sin xin (0,π2)\left( 0,\dfrac{\pi }{2} \right)

Explanation

Solution

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⁡xf(x)=\sin xin (0,π2)\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 considerf′(x){{f}^{'}}(x).
f′(x)=ddx(sin⁡x){{f}^{'}}(x)=\dfrac{d}{dx}\left( \sin x \right)
We know derivative of sin⁡x\sin x is cos⁡x\cos x , so the above equation becomes,
f′(x)=cos⁡x{{f}^{'}}(x)=\cos x
Now we will write f′(x)f'(x) in terms of sin⁡x\sin x.
We know, cos⁡x=sin⁡(90+x)\cos x=\sin (90+x), so the first derivative can be written as,
f′(x)=sin⁡(90+x)...........(i){{f}^{'}}(x)=\sin \left( 90+x \right)...........(i)
Now we will find the second derivative of the given expression.
So, let us considerf′(x){{f}^{'}}(x)and differentiating it will respect to ′x′'x', we get
f′′(x)=ddx(cos⁡x){{f}^{''}}(x)=\dfrac{d}{dx}\left( \cos x \right)
We know derivative of cos⁡x\cos x is −sin⁡x-\sin x , so the above equation becomes,
f′′(x)=−sin⁡x{{f}^{''}}(x)=-\sin x
We know, sin⁡(180+x)=−sin⁡x\sin (180+x)=-\sin x, so the second derivative can be written as,
f′′(x)=sin⁡(180+x){{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⁡(2×90+x)...........(ii){{f}^{''}}(x)=\sin \left( 2\times 90+x \right)...........(ii)
Now we will find the third derivative of the given expression.
So, let us considerf′′(x)f''(x)and differentiating it will respect to ′x′'x', we get
f′′′(x)=ddx(−sin⁡x){{f}^{'''}}(x)=\dfrac{d}{dx}\left( -\sin x \right)
We know derivative of sin⁡x\sin x is cos⁡x\cos x , so the above equation becomes,
f′′′(x)=−cos⁡x{{f}^{'''}}(x)=-\cos x
We know, sin⁡(270+x)=−cos⁡x\sin (270+x)=-\cos x, so the third derivative can be written as,
f′′′(x)=sin⁡(270+x){{f}^{'''}}(x)=\sin \left( 270+x \right)
Now we find the general term derivative we will try to write ′270′'270' in terms of ′90′'90', we get
f′′(x)=sin⁡(3×90+x)..........(iii){{f}^{''}}(x)=\sin \left( 3\times 90+x \right)..........(iii)
Similarly, we will find the fourth derivative, we get
So, let us considerf′′′(x)f'''(x)and differentiating it will respect to ′x′'x', we get
fIV(x)=ddx(−cos⁡x){{f}^{IV}}(x)=\dfrac{d}{dx}\left( -\cos x \right)
We know derivative of cos⁡x\cos x is −sin⁡x-\sin x , so the above equation becomes,
fIV(x)=−(−sin⁡x)=sin⁡x{{f}^{IV}}(x)=-(-\sin x)=\sin x
We know, sin⁡(360+x)=sin⁡x\sin (360+x)=\sin x, so the fourth derivative can be written as,
fIV(x)=sin⁡(360+x){{f}^{IV}}(x)=\sin \left( 360+x \right)
Now we find the general term derivative we will try to write ′360′'360' in terms of ′90′'90', we get
fIV(x)=sin⁡(4×90+x)..........(iv){{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⁡(1×90+x){{f}^{'}}(x)=\sin \left( 1\times 90+x \right)
f′′(x)=sin⁡(2×90+x){{f}^{''}}(x)=\sin \left( 2\times 90+x \right)
f′′′(x)=sin⁡(3×90+x){{f}^{'''}}(x)=\sin \left( 3\times 90+x \right)
fIV(x)=sin⁡(4×90+x){{f}^{IV}}(x)=\sin \left( 4\times 90+x \right)
By observing the above four equations we can write the nth derivative as,
fn(x)=sin⁡(n×90+x){{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\sin x function in the given range (0,π2)\left( 0,\dfrac{\pi }{2} \right). In this way we can observe the similarity between the derivatives in order to find nth derivative.