What is the derivative of 1/(sqrt(3x+5)) ?

The derivative of 1/(sqrt(3x+5)) is -3/(2(3x+5)^{3/2)}

What is the first derivative of 1/(sqrt(3x+5)) ?

The first derivative of 1/(sqrt(3x+5)) is -3/(2(3x+5)^{3/2)}

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\frac{d}{d x} (\frac{1}{3 x + 5})

- \frac{3}{2 ( 3 x + 5 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{1}{3 x + 5})

Apply exponent rule:

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

= \frac{d}{d x} ((3 x + 5)^{- \frac{1}{2}})

Apply the chain rule:

- \frac{1}{2 ( 3 x + 5 ) ^{\frac{3}{2}}} \frac{d}{d x} (3 x + 5)

= - \frac{1}{2 ( 3 x + 5 ) ^{\frac{3}{2}}} \frac{d}{d x} (3 x + 5)

\frac{d}{d x} (3 x + 5) = 3

= - \frac{1}{2 ( 3 x + 5 ) ^{\frac{3}{2}}} \cdot 3

Simplify

- \frac{1}{2 ( 3 x + 5 ) ^{\frac{3}{2}}} \cdot 3 : - \frac{3}{2 ( 3 x + 5 ) ^{\frac{3}{2}}}

= - \frac{3}{2 ( 3 x + 5 ) ^{\frac{3}{2}}}

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

\frac{d}{d x} ([x + (x + sin^{2} (x))^{7}]^{4})

\frac{\partial}{\partial x} (cos^{3} (x^{3} y^{8}))

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 e^{- 2 t}

(1 + \frac{3}{y} + x) \frac{d y}{d x} + y = \frac{3}{x} - 1

What is the derivative of 1/(sqrt(3x+5)) ?
The derivative of 1/(sqrt(3x+5)) is -3/(2(3x+5)^{3/2)}
What is the first derivative of 1/(sqrt(3x+5)) ?
The first derivative of 1/(sqrt(3x+5)) is -3/(2(3x+5)^{3/2)}