Question

Find a formula for the nth derivative of \[\sin \left( {ax + b} \right)\] and \[\cos \left( {ax + b} \right)\].

Answer 1

Hint: In order to solve this question, for first part we will first differentiate the given function using \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] and for second part by using \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\] . After that we will differentiate again and find a few more derivatives and try to transform in the form of a given function and in the form of the first derivative. Then finally generalise them to get the nth derivative.

Complete step by step answer:
For first part, the given function is,
\[f\left( x \right) = \sin \left( {ax + b{\text{ }}} \right){\text{ }} - - - \left( i \right)\]
Now in order to find the nth derivative, we will find the first few derivatives. For this we shall consider 1st, 2nd, and 3rd derivatives of the given function.Now we know that
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
Therefore, on differentiating equation \[\left( i \right)\] we get
\[{f^{\left( 1 \right)}}\left( x \right) = \cos \left( {ax + b} \right) \times \dfrac{d}{{dx}}\left( {ax + b} \right)\]
\[ \Rightarrow {f^{\left( 1 \right)}}\left( x \right) = \cos \left( {ax + b} \right) \times a\]
Now we know \[\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)\]
\[ \Rightarrow {f^{\left( 1 \right)}}\left( x \right) = a\sin \left( {ax + b + \dfrac{\pi }{2}} \right){\text{ }} - - - \left( {ii} \right)\]

Now let us consider \[{f^{\left( 1 \right)}}\left( x \right)\] and differentiating it, we will get
\[{f^{\left( 2 \right)}}\left( x \right) = \dfrac{d}{{dx}}\left( {a\sin \left( {ax + b + \dfrac{\pi }{2}} \right)} \right) \times \dfrac{d}{{dx}}\left( {ax + b + \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = a\cos \left( {ax + b + \dfrac{\pi }{2}} \right) \times a\]
Therefore, on solving we get
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = {a^2}\cos \left( {ax + b + \dfrac{\pi }{2}} \right)\]
Now transforming in the form of first derivative by using \[\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)\] , we can write
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = {a^2}\sin \left( {ax + b + \dfrac{\pi }{2} + \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = {a^2}\sin \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right){\text{ }} - - - \left( {iii} \right)\]

Now similarly let us consider \[{f^{\left( 2 \right)}}\left( x \right)\] and differentiating it, we will get
\[{f^{\left( 3 \right)}}\left( x \right) = \dfrac{d}{{dx}}\left( {{a^2}\sin \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right)} \right) \times \dfrac{d}{{dx}}\left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = {a^2}\cos \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right) \times a\]
Therefore, on solving we have
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = {a^3}\cos \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right)\]

Now transforming in the form of first derivative by using \[\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)\] , we can write
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = {a^3}\sin \left( {ax + b + 2 \cdot \dfrac{\pi }{2} + \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = {a^3}\sin \left( {ax + b + 3 \cdot \dfrac{\pi }{2}} \right){\text{ }} - - - \left( {iv} \right)\]
and so on….
Therefore, by observing equation \[\left( {ii} \right),\left( {iii} \right)\] and \[\left( {iv} \right)\] we get the nth derivative as,
\[{f^{\left( n \right)}}\left( x \right) = {a^n}\sin \left( {ax + b + n \cdot \dfrac{\pi }{2}} \right)\]
which is the required answer.

Therefore, the formula for the nth derivative of \[\sin \left( {ax + b} \right)\] is $ {a^n}\sin \left( {ax + b + n \cdot \dfrac{\pi }{2}} \right)$.

Now we will do same for second part:
The given function is
\[f\left( x \right) = \cos \left( {ax + b{\text{ }}} \right){\text{ }} - - - \left( i \right)\]
Now in order to find the nth derivative, we will find the first few derivatives. For this we shall consider 1st, 2nd, and 3rd derivatives of the given function.
Now we know that
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]

Therefore, on differentiating equation \[\left( i \right)\] we get
\[{f^{\left( 1 \right)}}\left( x \right) = - \sin \left( {ax + b} \right) \times \dfrac{d}{{dx}}\left( {ax + b} \right)\]
\[ \Rightarrow {f^{\left( 1 \right)}}\left( x \right) = - \sin \left( {ax + b} \right) \times a\]
Now we know \[ - \sin \theta = \cos \left( {\dfrac{\pi }{2} + \theta } \right)\]
\[ \Rightarrow {f^{\left( 1 \right)}}\left( x \right) = a\cos \left( {ax + b + \dfrac{\pi }{2}} \right){\text{ }} - - - \left( {ii} \right)\]
Now let us consider \[{f^{\left( 1 \right)}}\left( x \right)\] and differentiating it, we will get
\[{f^{\left( 2 \right)}}\left( x \right) = \dfrac{d}{{dx}}\left( {a\cos \left( {ax + b + \dfrac{\pi }{2}} \right)} \right) \times \dfrac{d}{{dx}}\left( {ax + b + \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = - a\sin \left( {ax + b + \dfrac{\pi }{2}} \right) \times a\]

Therefore, on solving we get
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = - {a^2}\sin \left( {ax + b + \dfrac{\pi }{2}} \right)\]
Now transforming in the form of first derivative by using \[ - \sin \theta = \cos \left( {\dfrac{\pi }{2} + \theta } \right)\] , we can write
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = {a^2}\cos \left( {ax + b + \dfrac{\pi }{2} + \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 2 \right)}}\left( x \right) = {a^2}\cos \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right){\text{ }} - - - \left( {iii} \right)\]

Now similarly let us consider \[{f^{\left( 2 \right)}}\left( x \right)\] and differentiating it, we will get
\[{f^{\left( 3 \right)}}\left( x \right) = \dfrac{d}{{dx}}\left( {{a^2}\cos \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right)} \right) \times \dfrac{d}{{dx}}\left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = - {a^2}\sin \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right) \times a\]
Therefore, on solving we have
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = - {a^3}\sin \left( {ax + b + 2 \cdot \dfrac{\pi }{2}} \right)\]

Now transforming in the form of first derivative by using \[ - \sin \theta = \cos \left( {\dfrac{\pi }{2} + \theta } \right)\] , we can write
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = {a^3}\cos \left( {ax + b + 2 \cdot \dfrac{\pi }{2} + \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow {f^{\left( 3 \right)}}\left( x \right) = {a^3}\cos \left( {ax + b + 3 \cdot \dfrac{\pi }{2}} \right){\text{ }} - - - \left( {iv} \right)\]
and so on….
Therefore, by observing equation \[\left( {ii} \right),\left( {iii} \right)\] and \[\left( {iv} \right)\] we get the nth derivative as,
\[{f^{\left( n \right)}}\left( x \right) = {a^n}\cos \left( {ax + b + n \cdot \dfrac{\pi }{2}} \right)\]

Therefore, the formula for the nth derivative of \[\cos \left( {ax + b} \right)\] is ${a^n}\cos \left( {ax + b + n \cdot \dfrac{\pi }{2}} \right)$.

Note: In this type of questions, the key point is that we have to simplify at least 3-4 differentiation of the given function to get some kind of variation. Also note that the main key point is expressing all the derivatives in the form of a given function. In this way we can observe the similarity between the derivatives in order to find the nth derivative.