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

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

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

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

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

- \frac{5 ( 6 x + 1 )}{( 3 x ^{2} + x + 2 ) ^{6}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= - \frac{5}{( 3 x ^{2} + x + 2 ) ^{6}} (6 x + 1)

Multiply fractions:

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

= - \frac{5 ( 6 x + 1 )}{( 3 x ^{2} + x + 2 ) ^{6}}

\int \frac{- 10}{( x - 2 ) ( x ^{2} + 4 )} d x

\int \frac{tan ^{4} ( x )}{sec ^{5} ( x )} d x

y^{^{'}} = 4 y^{2} x

\int 13 tan^{5} (x) d x

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

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