What is the derivative of (ax^3+bx+c/(dx^3-e)) ?

The derivative of (ax^3+bx+c/(dx^3-e)) is (-2bdx^3-3eax^2-3cdx^2-eb)/((dx^3-e)^2)

What is the first derivative of (ax^3+bx+c/(dx^3-e)) ?

The first derivative of (ax^3+bx+c/(dx^3-e)) is (-2bdx^3-3eax^2-3cdx^2-eb)/((dx^3-e)^2)

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\frac{d}{d x} (\frac{a x ^{3} + b x + c}{d x ^{3} - e})

\frac{- 2 b d x ^{3} - 3 e a x ^{2} - 3 c d x ^{2} - e b}{( d x ^{3} - e ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{a x ^{3} + b x + c}{d x ^{3} - e})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( a x ^{3} + b x + c ) ( d x ^{3} - e ) - \frac{d}{d x} ( d x ^{3} - e ) ( a x ^{3} + b x + c )}{( d x ^{3} - e ) ^{2}}

\frac{d}{d x} (a x^{3} + b x + c) = 3 a x^{2} + b

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

= \frac{( 3 a x ^{2} + b ) ( d x ^{3} - e ) - 3 d x ^{2} ( a x ^{3} + b x + c )}{( d x ^{3} - e ) ^{2}}

Expand

(3 a x^{2} + b) (d x^{3} - e) - 3 d x^{2} (a x^{3} + b x + c) : - 2 b d x^{3} - 3 e a x^{2} - 3 c d x^{2} - e b

= \frac{- 2 b d x ^{3} - 3 e a x ^{2} - 3 c d x ^{2} - e b}{( d x ^{3} - e ) ^{2}}

\int_{0}^{1} \frac{6 + x}{1 + x ^{2}} d x

f (x) = \frac{x ^{3}}{x + 1}, at x = 1

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

y = (x - 1)^{3}

\int tan (x) ln (cos (x)) d x

What is the derivative of (ax^3+bx+c/(dx^3-e)) ?
The derivative of (ax^3+bx+c/(dx^3-e)) is (-2bdx^3-3eax^2-3cdx^2-eb)/((dx^3-e)^2)
What is the first derivative of (ax^3+bx+c/(dx^3-e)) ?
The first derivative of (ax^3+bx+c/(dx^3-e)) is (-2bdx^3-3eax^2-3cdx^2-eb)/((dx^3-e)^2)