What is the derivative of 1/((3x^3+5x+7^{3/4)}) ?

The derivative of 1/((3x^3+5x+7^{3/4)}) is -(3(9x^2+5))/(4(3x^3+5x+7)^{7/4)}

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

The first derivative of 1/((3x^3+5x+7^{3/4)}) is -(3(9x^2+5))/(4(3x^3+5x+7)^{7/4)}

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\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 7 ) ^{\frac{3}{4}}})

- \frac{3 ( 9 x ^{2} + 5 )}{4 ( 3 x ^{3} + 5 x + 7 ) ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 7 ) ^{\frac{3}{4}}})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{3} + 5 x + 7) = 9 x^{2} + 5

= - \frac{3}{4 ( 3 x ^{3} + 5 x + 7 ) ^{\frac{7}{4}}} (9 x^{2} + 5)

Multiply fractions:

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

= - \frac{3 ( 9 x ^{2} + 5 )}{4 ( 3 x ^{3} + 5 x + 7 ) ^{\frac{7}{4}}}

(e^{y} + 6 \cdot x) \cdot d x + (x \cdot e^{y}) \cdot d y = 0

x \to 16 lim (4 - x)

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

y = arcsin (6 x + 1)

\int \frac{2 x ^{2} + x - 8}{x ^{3} + 4 x} d x

What is the derivative of 1/((3x^3+5x+7^{3/4)}) ?
The derivative of 1/((3x^3+5x+7^{3/4)}) is -(3(9x^2+5))/(4(3x^3+5x+7)^{7/4)}
What is the first derivative of 1/((3x^3+5x+7^{3/4)}) ?
The first derivative of 1/((3x^3+5x+7^{3/4)}) is -(3(9x^2+5))/(4(3x^3+5x+7)^{7/4)}