What is the derivative of 1/(a+b*(x/d ^c)) ?

The derivative of 1/(a+b*(x/d ^c)) is -(bcd^cx^{c-1})/((ad^c+bx^c)^2)

What is the first derivative of 1/(a+b*(x/d ^c)) ?

The first derivative of 1/(a+b*(x/d ^c)) is -(bcd^cx^{c-1})/((ad^c+bx^c)^2)

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\frac{d}{d x} (\frac{1}{a + b \cdot ( \frac{x}{d} ) ^{c}})

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

Solution steps

\frac{d}{d x} (\frac{1}{a + b ( \frac{x}{d} ) ^{c}})

Treat

a, b, d, c

as constants

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d t} + 5 y = 80, y (1) = 100

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

\int \frac{1}{tan ( x ) sec ( x )} d x

\frac{\partial}{\partial x} (\frac{y ^{2}}{2 x})

\int_{- 1}^{1} - x^{2} + 1 d x

What is the derivative of 1/(a+b*(x/d ^c)) ?
The derivative of 1/(a+b*(x/d ^c)) is -(bcd^cx^{c-1})/((ad^c+bx^c)^2)
What is the first derivative of 1/(a+b*(x/d ^c)) ?
The first derivative of 1/(a+b*(x/d ^c)) is -(bcd^cx^{c-1})/((ad^c+bx^c)^2)