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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

x \cdot sec (x)

\int \frac{1}{x 16 x ^{2} - 9} d x

\frac{\partial}{\partial x} (\frac{7}{x ^{3}})

\frac{d}{d x} (\frac{π x}{2})

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