What is the derivative of 1/(q+ke^q) ?

The derivative of 1/(q+ke^q) is -(1+ke^q)/((q+ke^q)^2)

What is the first derivative of 1/(q+ke^q) ?

The first derivative of 1/(q+ke^q) is -(1+ke^q)/((q+ke^q)^2)

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derivative of

\frac{1}{q + k e ^{q}}

- \frac{1 + k e ^{q}}{( q + k e ^{q} ) ^{2}}

Solution steps

\frac{d}{d q} (\frac{1}{q + k e ^{q}})

Apply exponent rule:

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

= \frac{d}{d q} ((q + k e^{q})^{- 1})

Apply the chain rule:

- \frac{1}{( q + k e ^{q} ) ^{2}} \frac{d}{d q} (q + k e^{q})

= - \frac{1}{( q + k e ^{q} ) ^{2}} \frac{d}{d q} (q + k e^{q})

\frac{d}{d q} (q + k e^{q}) = 1 + k e^{q}

= - \frac{1}{( q + k e ^{q} ) ^{2}} (1 + k e^{q})

Simplify

- \frac{1}{( q + k e ^{q} ) ^{2}} (1 + k e^{q}) : - \frac{1 + k e ^{q}}{( q + k e ^{q} ) ^{2}}

= - \frac{1 + k e ^{q}}{( q + k e ^{q} ) ^{2}}

\frac{\partial}{\partial y} (4 x^{\frac{y}{z}})

y^{^{''}} + y = 3 e^{- 8 x}

\int_{0}^{4} x^{3} 7 - x^{2} d x

f (x) = x^{3} - \frac{2}{x ^{2}} + 2, (1, 3)

\frac{\partial}{\partial x} (x^{1})

What is the derivative of 1/(q+ke^q) ?
The derivative of 1/(q+ke^q) is -(1+ke^q)/((q+ke^q)^2)
What is the first derivative of 1/(q+ke^q) ?
The first derivative of 1/(q+ke^q) is -(1+ke^q)/((q+ke^q)^2)