Calc2.61

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Let f(x)=a(x)\cdot b(x)

Then f(x+\Delta x)=a(x+\Delta x)\cdot b(x+\Delta x)

So {\frac  {f(x+\Delta x)-f(x)}{\Delta x}}={\frac  {a(x+\Delta x)b(x+\Delta x)-a(x)b(x)}{\Delta x}}={\frac  {a(x+\Delta x)b(x+\Delta x)+a(x+\Delta x)b(x)-a(x+\Delta x)b(x)-a(x)b(x)}{\Delta x}}

={\frac  {a(x+\Delta x)b(x+\Delta x)+a(x+\Delta x)b(x)}{\Delta x}}-{\frac  {a(x+\Delta x)b(x)-a(x)b(x)}{\Delta x}}

=a(x+\Delta x)\cdot {\frac  {b(x+\Delta x)+b(x)}{\Delta x}}-{\frac  {a(x+\Delta x)+a(x)}{\Delta x}}\cdot b(x)

Taking the limit: \lim _{{\Delta x\to 0}}{\frac  {f(x+\Delta x)-f(x)}{\Delta x}}=\lim _{{\Delta x\to 0}}\left[a(x+\Delta x)\cdot {\frac  {b(x+\Delta x)+b(x)}{\Delta x}}\right]-\lim _{{\Delta x\to 0}}\left[{\frac  {a(x+\Delta x)+a(x)}{\Delta x}}\cdot b(x)\right]

Hence f'(x)=a(x)\cdot b'(x)+a'(x)\cdot b(x)


Note: It can be shown that \lim _{{y\to c}}g(y)\cdot h(y)=g(c)h(c)\, if \lim _{{y\to c}}g(y)=g(c) and \lim _{{y\to c}}h(y)=h(c)