What is the derivative of 4/(2x+5) ?

The derivative of 4/(2x+5) is -8/((2x+5)^2)

What is the first derivative of 4/(2x+5) ?

The first derivative of 4/(2x+5) is -8/((2x+5)^2)

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\frac{d}{d x} (\frac{4}{2 x + 5})

- \frac{8}{( 2 x + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4}{2 x + 5})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (\frac{1}{2 x + 5})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (2 x + 5) = 2

= 4 (- \frac{1}{( 2 x + 5 ) ^{2}} \cdot 2)

Simplify

4 (- \frac{1}{( 2 x + 5 ) ^{2}} \cdot 2) : - \frac{8}{( 2 x + 5 ) ^{2}}

= - \frac{8}{( 2 x + 5 ) ^{2}}

ln (\frac{x - 5}{x - 4})

\frac{d}{d x} (- x e^{- \frac{x ^{2}}{2}})

f^{^{'}} (x) = 2 sec (x) + 3 tan (x) - tan (\frac{π}{3})

y^{^{''}} + (\frac{1}{x}) y^{^{'}} = 0

\int 4 e^{2} d x

What is the derivative of 4/(2x+5) ?
The derivative of 4/(2x+5) is -8/((2x+5)^2)
What is the first derivative of 4/(2x+5) ?
The first derivative of 4/(2x+5) is -8/((2x+5)^2)