Proof of the Product Rule in Differentiation

 Product Rule Proof:

let 

\[y(x) = u(x) \cdot v(x) \]

Taking the natural log on both sides

\[ \ln y(x) = \ln (u(x)\cdot v(x)) \]

\[ \ln y(x) = \ln u(x) + \ln v(x) \]

Taking the Differential

\[\dfrac{1}{y(x)} y'(x) = \dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x) \]

\[ y'(x) = y(x) ( \dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x)) \]

But \(y(x) = u(x) \cdot v(x)\)

\[ y'(x) = u(x) \cdot v(x) (\dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x)) \]

\[ y'(x) = v(x) \cdot u'(x) + u(x) \cdot v'(x)) \]


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