What is the derivative of y=(x^3+8x^2)^{-1} ?

The derivative of y=(x^3+8x^2)^{-1} is -(3x+16)/(x^3(x+8)^2)

What is the first derivative of y=(x^3+8x^2)^{-1} ?

The first derivative of y=(x^3+8x^2)^{-1} is -(3x+16)/(x^3(x+8)^2)

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derivative of

y = (x^{3} + 8 x^{2})^{- 1}

- \frac{3 x + 16}{x ^{3} ( x + 8 ) ^{2}}

Solution steps

\frac{d}{d x} ((x^{3} + 8 x^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{3} + 8 x ^{2} ) ^{2}} \frac{d}{d x} (x^{3} + 8 x^{2})

= - \frac{1}{( x ^{3} + 8 x ^{2} ) ^{2}} \frac{d}{d x} (x^{3} + 8 x^{2})

\frac{d}{d x} (x^{3} + 8 x^{2}) = 3 x^{2} + 16 x

= - \frac{1}{( x ^{3} + 8 x ^{2} ) ^{2}} (3 x^{2} + 16 x)

Simplify

- \frac{1}{( x ^{3} + 8 x ^{2} ) ^{2}} (3 x^{2} + 16 x) : - \frac{3 x + 16}{x ^{3} ( x + 8 ) ^{2}}

= - \frac{3 x + 16}{x ^{3} ( x + 8 ) ^{2}}

What is the derivative of y=(x^3+8x^2)^{-1} ?
The derivative of y=(x^3+8x^2)^{-1} is -(3x+16)/(x^3(x+8)^2)
What is the first derivative of y=(x^3+8x^2)^{-1} ?
The first derivative of y=(x^3+8x^2)^{-1} is -(3x+16)/(x^3(x+8)^2)