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bittusohan · May 12, 2012, 08:52 PM

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

ebaines · May 14, 2012, 06:08 AM

Originally Posted by bittusohan

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.