Proof of Quotient Rule in Differentiation

 Proof of Quotient Rule:

Let 

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

Taking the natural Logarithm

\[\ln y(x) = \ln \dfrac{u(x)}{v(x)}\]

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

Taking the Differential 

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

Multiplying by y(x)

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

But \( y(x) = \dfrac{u(x)}{v(x)} \)

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

\[ y'(x) = \dfrac{u'(x)}{v(x)} - \dfrac{u(x)v'(x)}{v(x)^2} \]

\[ y'(x) = \dfrac{u'(x)v(x) - u'(x)v'(x)}{v(x)^2} \]

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