Answer:
Step-by-step explanation:
We have,
[tex]\tt{y=sin(5x)\,cos(4x)}[/tex]
[tex]\tt{\implies\,y=\dfrac{1}{2}\cdot2\,sin(5x)\,cos(4x)}[/tex]
[tex]\tt{\implies\,y=\dfrac{1}{2}\left\{sin\left(5x+4x\right)+sin\left(5x-4x\right)\right\}}[/tex]
[tex]\tt{\implies\,y=\dfrac{1}{2}\left\{sin\left(9x\right)+sin\left(x\right)\right\}}[/tex]
[tex]\tt{\implies\,y=\dfrac{1}{2}\,sin(9x)+\dfrac{1}{2}\,sin(x)}[/tex]
We know,
[tex]\boxed{\bf{\dfrac{{d}^{n}}{d{x}^{n}}\Big(sin(ax+b)\Big)={a}^{n}\cdot\,sin\left(ax+b+\dfrac{n\pi}{2}\right)}}[/tex]
So,
[tex]\tt{\implies\,y_{n}=\dfrac{1}{2}\cdot{9}^{n}\,sin\left(9x+\dfrac{n\pi}{2}\right)+\dfrac{1}{2}\,sin\left(x+\dfrac{n\pi}{2}\right)}[/tex]
[tex]\tt{\implies\,y_{n}=\dfrac{1}{2}\left\{{9}^{n}\cdot\,sin\left(9x+\dfrac{n\pi}{2}\right)+sin\left(x+\dfrac{n\pi}{2}\right)\right\}}[/tex]