What is the derivative of h(x)= 1/((-3x^2+3x-2)^3) ?

The derivative of h(x)= 1/((-3x^2+3x-2)^3) is -(3(-6x+3))/((-3x^2+3x-2)^4)

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

The first derivative of h(x)= 1/((-3x^2+3x-2)^3) is -(3(-6x+3))/((-3x^2+3x-2)^4)

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Multiply fractions:

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

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

126 x (9 x^{2} + 6)^{6}

\int 20 tan^{3} (x) d x

\int_{- 1}^{0} x + 1 d x

y = x + 1, y = 7 x - 17, y = - 2 x - 2

y = ∣ 9 x ∣, y = x^{2} - 10

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