[tex]Answer: \frac{\partial y}{\partial x}=8x^{8sinx}(cosx.logx+\frac{sinx}{x})[/tex]
Step-by-step explanation:
Since we have given that
[tex]y=x^{8sinx}[/tex]
By using logarithmic on both sides we get,
[tex]log y= 8 sinx. logx[/tex]
(∵ [tex]log(a^m)=m.loga[/tex])
Now, differentiating on both sides ,we get,
[tex]\frac{1}{y}.\frac{\partial y}{\partial x}=8(\frac{\partial sinx}{\partial x}.logx+sinx \frac{\partial logx}{\partial x})[/tex]
[tex]\\\frac{1}{y}\frac{\partial y}{\partial x}=8(cosx.logx+sinx.\frac{1}{x})[/tex]
[tex]\\\frac{\partial y}{\partial y}=8y(cosx.logx+\frac{sinx}{x})\\\\\frac{\partial y}{\partial x}=8x^{8sinx}(cosx.logx+\frac{sinx}{x})[/tex]