What is the derivative of 1/(1+1/4 x) ?

The derivative of 1/(1+1/4 x) is -4/((4+x)^2)

What is the first derivative of 1/(1+1/4 x) ?

The first derivative of 1/(1+1/4 x) is -4/((4+x)^2)

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\frac{d}{d x} (\frac{1}{1 + \frac{1}{4} x})

- \frac{4}{( 4 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{1 + \frac{1}{4} x})

Apply exponent rule:

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

= \frac{d}{d x} ((1 + \frac{1}{4} x)^{- 1})

Apply the chain rule:

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

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

\frac{d}{d x} (1 + \frac{1}{4} x) = \frac{1}{4}

= - \frac{1}{( 1 + \frac{1}{4} x ) ^{2}} \cdot \frac{1}{4}

Simplify

- \frac{1}{( 1 + \frac{1}{4} x ) ^{2}} \cdot \frac{1}{4} : - \frac{4}{( 4 + x ) ^{2}}

= - \frac{4}{( 4 + x ) ^{2}}

\int (a x^{4} + b x^{3} + 3 c) d x

\frac{\partial}{\partial x} (\frac{( 2 π \cdot l ^{2} )}{x})

\int_{0}^{16} x d x

\int_{0}^{4} 4 y^{2} - \frac{y ^{4}}{4} d y

x \to a lim (f (x) + g (x))

What is the derivative of 1/(1+1/4 x) ?
The derivative of 1/(1+1/4 x) is -4/((4+x)^2)
What is the first derivative of 1/(1+1/4 x) ?
The first derivative of 1/(1+1/4 x) is -4/((4+x)^2)