What is the derivative of y=(x+6)/(x^3+x-7) ?

The derivative of y=(x+6)/(x^3+x-7) is (-2x^3-18x^2-13)/((x^3+x-7)^2)

What is the first derivative of y=(x+6)/(x^3+x-7) ?

The first derivative of y=(x+6)/(x^3+x-7) is (-2x^3-18x^2-13)/((x^3+x-7)^2)

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derivative of

y = \frac{x + 6}{x ^{3} + x - 7}

\frac{- 2 x ^{3} - 18 x ^{2} - 13}{( x ^{3} + x - 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 7})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x + 6 ) ( x ^{3} + x - 7 ) - \frac{d}{d x} ( x ^{3} + x - 7 ) ( x + 6 )}{( x ^{3} + x - 7 ) ^{2}}

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

\frac{d}{d x} (x^{3} + x - 7) = 3 x^{2} + 1

= \frac{1 \cdot ( x ^{3} + x - 7 ) - ( 3 x ^{2} + 1 ) ( x + 6 )}{( x ^{3} + x - 7 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{3} + x - 7 ) - ( 3 x ^{2} + 1 ) ( x + 6 )}{( x ^{3} + x - 7 ) ^{2}} : \frac{- 2 x ^{3} - 18 x ^{2} - 13}{( x ^{3} + x - 7 ) ^{2}}

= \frac{- 2 x ^{3} - 18 x ^{2} - 13}{( x ^{3} + x - 7 ) ^{2}}

(y^{4} - x^{2}) \frac{d y}{d x} = - x \cdot y

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0

\frac{\partial}{\partial x} (7 x ln (x y))

\frac{d}{d x} (1 3^{x})

\int_{2}^{4} \frac{1}{∣ x - 3 ∣ ^{\frac{1}{2}}} d x

What is the derivative of y=(x+6)/(x^3+x-7) ?
The derivative of y=(x+6)/(x^3+x-7) is (-2x^3-18x^2-13)/((x^3+x-7)^2)
What is the first derivative of y=(x+6)/(x^3+x-7) ?
The first derivative of y=(x+6)/(x^3+x-7) is (-2x^3-18x^2-13)/((x^3+x-7)^2)