What is the derivative of 1/((2x^3+3x+1)^{3/4)} ?

The derivative of 1/((2x^3+3x+1)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+1)^{7/4)}

What is the first derivative of 1/((2x^3+3x+1)^{3/4)} ?

The first derivative of 1/((2x^3+3x+1)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+1)^{7/4)}

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derivative of

\frac{1}{( 2 x ^{3} + 3 x + 1 ) ^{\frac{3}{4}}}

- \frac{3 ( 6 x ^{2} + 3 )}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{1}{( 2 x ^{3} + 3 x + 1 ) ^{\frac{3}{4}}})

Apply exponent rule:

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

= \frac{d}{d x} ((2 x^{3} + 3 x + 1)^{- \frac{3}{4}})

Apply the chain rule:

- \frac{3}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}} \frac{d}{d x} (2 x^{3} + 3 x + 1)

= - \frac{3}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}} \frac{d}{d x} (2 x^{3} + 3 x + 1)

\frac{d}{d x} (2 x^{3} + 3 x + 1) = 6 x^{2} + 3

= - \frac{3}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}} (6 x^{2} + 3)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{3 ( 6 x ^{2} + 3 )}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}}

y = e^{2 x}, at x = ln (2)

\frac{\partial}{\partial y} (y cos (x y) + e^{x})

\int cos^{2} (t) - sin^{2} (t) d t

\frac{\partial}{\partial x} (sin (x^{1} y^{3}))

What is the derivative of 1/((2x^3+3x+1)^{3/4)} ?
The derivative of 1/((2x^3+3x+1)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+1)^{7/4)}
What is the first derivative of 1/((2x^3+3x+1)^{3/4)} ?
The first derivative of 1/((2x^3+3x+1)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+1)^{7/4)}