Krishna
1
Proof We know the definition of derivative that is: \frac{d}{dx} f(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} [math]\frac{d}{dx}[ f(x)g(x)] = \lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h} [/math] I am Adding and subtracting with f(x+h)g(x) . There is no value change in the equation. [math]\frac{d}{dx}[ f(x)g(x)] = \lim_{h \to 0}\frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h} [/math] Extracting f(x+h) as the common from the first two terms and g(x) from last two [math]\frac{d}{dx}[ f(x)g(x)] = \lim_{h \to 0}\frac{f(x+h)[g(x+h) - g(x)] + g(x)[f(x+h) - f(x)]}{h} [/math] [math]\frac{d}{dx}[ f(x)g(x)] = \lim_{h \to 0}[f(x+h)]\lim_{h \to 0}\frac{[g(x+h) - g(x)]}{h} + \lim_{h \to 0}[g(x)]\lim_{h \to 0}\frac{[f(x+h) - f(x)]}{h} [/math] \lim_{h \to 0}f(x+h)= f(x) [math] \lim_{h \to 0}\frac{[g(x+h) - g(x)]}{h}[/math]=g'(x) \lim_{h \to 0} g(x)= g(x) [math]\lim_{h \to 0}\frac{[f(x+h) - f(x)]}{h}[/math]=f'(x) \frac{d}{dx} f(x)g(x)= f(x)g'(x)+f'(x)g'(x).
Sangeetha Pulapaka
1
If y = f(x).g(x) then we find the derivative of y by first finding out the derivative of both f(x) and g(x) We know the formula, \frac{d}{dx}f(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} Now find the product of two functions f(x) and g(x) by applying the definition of the derivative (shown above) [math]\frac{d}{dx}[f(x)g(x)][/math] = \lim_{h\to0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h} In the above step we add and subtract two similar terms in the middle is so that the value of the expression does not change. Then we take f(x+h) common and g(x) common to get \lim_{h\to 0}\Big( f(x+h)\frac{g(x+h)-g(x)}{h} +g(x)\frac{f(x+h)-f(x)}{h}\Big) =\lim_{h\to 0}\Big(f(x+h)\Big)\lim_{h\to 0}\Big(\frac{g(x+h)-g(x)}{h}\Big)+\lim_{h\to 0}\Big(g(x)\Big) \lim_{h\to 0}\Big(\frac{f(x+h)-f(x)}{h}\Big) = f(x).g'(x)+g(x) f'(x)