What is the derivative of 1/((2x^3+3x+5)^{3/4)} ?

The derivative of 1/((2x^3+3x+5)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+5)^{7/4)}

What is the first derivative of 1/((2x^3+3x+5)^{3/4)} ?

The first derivative of 1/((2x^3+3x+5)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+5)^{7/4)}

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derivative of

\frac{1}{( 2 x ^{3} + 3 x + 5 ) ^{\frac{3}{4}}}

- \frac{3 ( 6 x ^{2} + 3 )}{4 ( 2 x ^{3} + 3 x + 5 ) ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{1}{( 2 x ^{3} + 3 x + 5 ) ^{\frac{3}{4}}})

Apply exponent rule:

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

= \frac{d}{d x} ((2 x^{3} + 3 x + 5)^{- \frac{3}{4}})

Apply the chain rule:

- \frac{3}{4 ( 2 x ^{3} + 3 x + 5 ) ^{\frac{7}{4}}} \frac{d}{d x} (2 x^{3} + 3 x + 5)

= - \frac{3}{4 ( 2 x ^{3} + 3 x + 5 ) ^{\frac{7}{4}}} \frac{d}{d x} (2 x^{3} + 3 x + 5)

\frac{d}{d x} (2 x^{3} + 3 x + 5) = 6 x^{2} + 3

= - \frac{3}{4 ( 2 x ^{3} + 3 x + 5 ) ^{\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 + 5 ) ^{\frac{7}{4}}}

\int_{1}^{e} (ln (x))^{2} d x

\frac{d}{d x} (- sec (x))

\int \frac{x ^{2}}{x ^{3} + 1} d x

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

\int 8 x^{2} cos (2 x) d x

What is the derivative of 1/((2x^3+3x+5)^{3/4)} ?
The derivative of 1/((2x^3+3x+5)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+5)^{7/4)}
What is the first derivative of 1/((2x^3+3x+5)^{3/4)} ?
The first derivative of 1/((2x^3+3x+5)^{3/4)} is -(3(6x^2+3))/(4(2x^3+3x+5)^{7/4)}