# 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 $\lim _{{y\to c}}h(y)=h(c)$