The steps to prove the quotient rule of differentiation from the product rule of differentiation are presented along with examples, exercises and solutions.

Let function \( f(x) \) be given by the quotient of two functions \( u(x) \) and \( v(x) \) written as

\( \quad f(x) = \dfrac{u(x)}{ v(x)} \qquad (1) \)

Mutliply both sides of the above by \( v(x) \) and simplify to obtain

\( \quad f(x) v(x) = u(x) \)

Take the derivative of both sides of the above

\( \quad ( f(x) v(x))' = u'(x) \)

Apply product rule of differentiation to the left side

\( \quad f'(x) v(x) + f(x) v'(x) = u'(x) \)

Solve the above for \( f'(x) \)

\( \quad f'(x) = \dfrac{ u'(x) - f(x) v'(x) }{ v(x) }\)

Substitute \( f(x) \) by \( \dfrac{u(x)}{ v(x)} \) as given in \( (1) \) above

\( \quad f'(x) = \dfrac{ u'(x) - \dfrac{u(x)}{ v(x)} v'(x) }{ v(x) }\)

Rewrite the expression in the numerator with a common denominator and simplify to rewrite \( f'(x) \) as

\( \quad f'(x) = \dfrac{ u'(x) v(x) - u(x) v'(x) }{ (v(x))^2 }\)

Finally the quotient rule of differentiation

\[ \left(\dfrac{u}{v}\right)' = \dfrac{ u'(x) v(x) - u(x) v'(x) }{ (v(x))^2 } \qquad (I) \]

Find the derivatives of

a) \( \quad f(x) = \dfrac{x}{\ln x} \) b) \( \quad g(x) = \dfrac{\sin(x)}{\cos(x)} \)

Solution

a)

Let \( u(x) = x \) and \( v(x) = \ln x \) and write \( f(x) \) as the quotient of \( u \) and \( v \) as follows

\( \quad f(x) = \dfrac {u} {v} \)

Use the quotient rule in \( (I) \) to write the derivative \( f'(x) \) as

\( \quad f'(x) = \dfrac{u'(x) v(x) - u(x) v'(x) }{(v(x))^2} \qquad (2) \)

Given \( u(x) = x \) and \( v(x) = \ln x \), the derivatives \( u' \) and \( v' \) are given by

\( \quad u' = 1 \) and \( v' = \dfrac{1}{x} \)

Substitute \( u, u', v, v' \) by their expressions in \( (2) \) above to obtain

\( \quad f'(x) = \dfrac{ 1 \cdot \ln x - x \cdot \dfrac{1}{x} }{ \ln^2 x } \)

Simplify to obtain

\( \quad f'(x) = \dfrac{ \ln x - 1} { \ln^2 x } \)

b)

Let \( w(x) = \sin x \) and \( z(x) = \cos x \) and write \( g(x) \) as the quotient of \( w \) and \( z \) as follows

\( \quad f(x) = \dfrac {w} {z} \)

Use the quotient rule in \( (I) \) to write the derivative \( g'(x) \) as

\( \quad g'(x) = \dfrac{w'(x) z(x) - w(x) z'(x) }{(z(x))^2} \qquad (3) \)

Given \( w(x) = \sin x \) and \( z(x) = \cos x \), the derivatives \( w' \) and \( z' \)

\( \quad w' = \cos x \) and \( z' = - \sin x \)

Substitute \( w, w', z, z' \) by their expressions in \( (3) \) above to obtain

\( \quad g'(x) = \dfrac{ \cos x \cos x - \sin x (- \sin x) }{ \cos^2 x } \)

The numerator of the above is equal to \( \cos x \cos x - \sin x (- \sin x) = \cos^2 x + \sin^2 x = 1\), hence simplify to obtain

\( \quad g'(x) = \dfrac{ 1 }{ \cos^2 x } \)

Use the identity \( \sec x = \dfrac{1}{\cos x} \) to write the final answer as

\[ g'(x) = \sec^2 x \]

Calculate the derivative of \( h(x) = \dfrac{x \; \ln x}{\sin x \; e^x} \)

Solution

Let \( u(x) = x \; \ln x \) and \( v(x) = \sin x \; e^x \) and write \( h(x) \) as the quotient

\( h(x) = \dfrac{u(x)}{v(x)} \)

Use the quotient rule in \( (I) \) to write the derivative \( h'(x) \) as

\( \quad h'(x) = \dfrac{ u'(x) v(x) - u(x) v'(x) }{ (v(x))^2 } \qquad (4) \)

Given \( u(x) = x \; \ln x \) which is the product of two functions, the derivative \( u' \) is calculated using the product rule of differentiation

\( u'(x) = (x)'\ln x + x (\ln x)' = \ln x + x \dfrac{1}{x} = \ln x + 1 \)

\( v(x) = \sin x \; e^x \) is also the product of two functions, the derivative \( v' \) is calculated using the product rule of differentiation

\( v'(x) = (\sin x)'e^x + \sin x (e^x)' = \cos x e^x + \sin x e^x = (\cos x + \sin x) e^x \)

Substitute \( u, u', v, v' \) by their expressions in \( (4) \) above to obtain

\( \quad h'(x) = \dfrac{ (\ln x + 1) (\sin x \; e^x) - (x \; \ln x )((\cos x + \sin x) e^x) }{ (\sin x \; e^x )^2 } \)

which may be rewritten as

\[ \quad h'(x) = \dfrac{ (\ln x + 1) \sin x - x \; \ln x \; (\cos x + \sin x) }{ \sin^2 x \; e^{x} } \]

Find the derivatives of the functions

- \( \quad f(x) = \dfrac{\cos(x)}{\sin x} \)

- \( \quad g(x) = \dfrac{-2x+4}{e^x} \)

- \( \quad h(x) = \dfrac{ 2x \; e^x}{ x \sin x} \)

- \( \quad f'(x) = \dfrac{-\sin x \sin x - \cos x \cos x}{\sin^2 x} = \dfrac{-(\sin^2 x + \cos^ x)}{\sin^2 x} = \dfrac{-1}{\sin^2 x} = - \csc^2 x \)

- \( \quad g'(x) = \dfrac{-2 \; e^x - (-2x+4) \; e^x}{ (e^x)^2} = -\dfrac{2\left(-x+3\right)}{e^x} \)

- \( \quad h(x) = \dfrac{ (2(1+x)\; e^x) (x \sin x) - (2x \; e^x) ( \sin x + x \; \cos x)}{ (x \; \sin x)^2} \\ \qquad \qquad = \dfrac{ 2 x \sin x \; e^x + 2 x^2 \sin x \; e^x - 2 x \sin x \; e^x - 2 x^2 \cos x \; e^x} { (x \; \sin x)^2} \\ \qquad \qquad = 2\left(\csc \left(x\right)-\cot \left(x\right)\csc \left(x\right)\right) e^x \)