What is (\partial)/(\partial x)((c-b)/(a-x)) ?

The answer to (\partial)/(\partial x)((c-b)/(a-x)) is (c-b)/((a-x)^2)

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(\partial)/(\partial x)((c-b)/(a-x))

\frac{\partial}{\partial x} (\frac{c - b}{a - x})

\frac{c - b}{( a - x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{c - b}{a - x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= (c - b) \frac{\partial}{\partial x} (\frac{1}{a - x})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= (c - b) \frac{\partial}{\partial x} ((a - x)^{- 1})

Apply the chain rule:

- \frac{1}{( a - x ) ^{2}} \frac{\partial}{\partial x} (a - x)

= - \frac{1}{( a - x ) ^{2}} \frac{\partial}{\partial x} (a - x)

\frac{\partial}{\partial x} (a - x) = - 1

= (c - b) (- \frac{1}{( a - x ) ^{2}} (- 1))

Simplify

(c - b) (- \frac{1}{( a - x ) ^{2}} (- 1)) : \frac{c - b}{( a - x ) ^{2}}

= \frac{c - b}{( a - x ) ^{2}}

\frac{d}{d x} (m x^{n} - n x^{m})

y^{^{'}} = \frac{( 1 + x )}{x y ^{14}}

y = (x^{2} - 15)^{4}, at x = 4

\frac{d}{d x} (ln (- cos (x^{2}) + 1))

\frac{d}{d x} (sin (\frac{x}{x + 1}))

What is (\partial)/(\partial x)((c-b)/(a-x)) ?
The answer to (\partial)/(\partial x)((c-b)/(a-x)) is (c-b)/((a-x)^2)