How to find the derivative of xsinx by using first principle ?

How to find the derivative of xsinx by using first principle ?

I'll get you started. Using the definition of the derivative:


\frac {df(x)} {dx} = \lim _{h \to 0} \frac {f(x+h) - f(x)} h


For f(x) = xsinx this gives:


\frac {df(x)} {dx} = \lim _{h \to 0} \frac {(x+h)\sin(x+h) - x \sin(x)} h \\
= \lim _{h \to 0} \frac {(x+h)(\sin x \cos h + \cos x \sin h) - x \sin x} h \\
= \lim _{h \to 0} \frac {x \sin x \cos h + x \cos x \sin h + h \sin x \cos h + h \cos x \sin h - x \sin x} h


Recall that

\lim _{h \to 0} \cos(h) = 1\\
\lim _{h \to 0}\frac { \sin(h)} h = 1\\


From here you should be able to finish it out. Post back if you're still having difficulties.