What is the derivative of ((6+7x))/((x^4+5x-5)) ?

The derivative of ((6+7x))/((x^4+5x-5)) is (-21x^4-24x^3-65)/((x^4+5x-5)^2)

What is the first derivative of ((6+7x))/((x^4+5x-5)) ?

The first derivative of ((6+7x))/((x^4+5x-5)) is (-21x^4-24x^3-65)/((x^4+5x-5)^2)

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derivative of

\frac{( 6 + 7 x )}{( x ^{4} + 5 x - 5 )}

\frac{- 21 x ^{4} - 24 x ^{3} - 65}{( x ^{4} + 5 x - 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 + 7 x}{x ^{4} + 5 x - 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 + 7 x ) ( x ^{4} + 5 x - 5 ) - \frac{d}{d x} ( x ^{4} + 5 x - 5 ) ( 6 + 7 x )}{( x ^{4} + 5 x - 5 ) ^{2}}

\frac{d}{d x} (6 + 7 x) = 7

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

= \frac{7 ( x ^{4} + 5 x - 5 ) - ( 4 x ^{3} + 5 ) ( 6 + 7 x )}{( x ^{4} + 5 x - 5 ) ^{2}}

Expand

7 (x^{4} + 5 x - 5) - (4 x^{3} + 5) (6 + 7 x) : - 21 x^{4} - 24 x^{3} - 65

= \frac{- 21 x ^{4} - 24 x ^{3} - 65}{( x ^{4} + 5 x - 5 ) ^{2}}

k = 2 \sum \infty \frac{8}{4 ^{2 k}}

f (x) = - x + 3 sin (x)

\int 6 x 1 - 3 x^{2} d x

\frac{d}{d x} (- \frac{4}{x ^{4}})

\frac{\partial}{\partial x} (3 y^{2} e^{x} - 4 x)

What is the derivative of ((6+7x))/((x^4+5x-5)) ?
The derivative of ((6+7x))/((x^4+5x-5)) is (-21x^4-24x^3-65)/((x^4+5x-5)^2)
What is the first derivative of ((6+7x))/((x^4+5x-5)) ?
The first derivative of ((6+7x))/((x^4+5x-5)) is (-21x^4-24x^3-65)/((x^4+5x-5)^2)