What is (\partial)/(\partial x)(1/(1-bx)) ?

The answer to (\partial)/(\partial x)(1/(1-bx)) is b/((1-bx)^2)

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(\partial)/(\partial x)(1/(1-bx))

\frac{\partial}{\partial x} (\frac{1}{1 - b x})

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

Solution steps

\frac{\partial}{\partial x} (\frac{1}{1 - b x})

Treat

b

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{arctan ( 5 x )}{1 + 25 x ^{2}} d x

\frac{d}{d x} ((2 x - 7)^{4} (x^{2} + x + 1)^{5})

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

x \to \infty lim (- \frac{sin ^{2} ( x ) + 2}{2 x - 5})

\frac{\partial}{\partial x} (sin (3 x) cos (6 y))

What is (\partial)/(\partial x)(1/(1-bx)) ?
The answer to (\partial)/(\partial x)(1/(1-bx)) is b/((1-bx)^2)