What is the derivative of (x^4-3x^2+1/((2x+3)^4)) ?

The derivative of (x^4-3x^2+1/((2x+3)^4)) is (2(6x^3+6x^2-9x-4))/((2x+3)^5)

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

The first derivative of (x^4-3x^2+1/((2x+3)^4)) is (2(6x^3+6x^2-9x-4))/((2x+3)^5)

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\frac{d}{d x} (\frac{x ^{4} - 3 x ^{2} + 1}{( 2 x + 3 ) ^{4}})

\frac{2 ( 6 x ^{3} + 6 x ^{2} - 9 x - 4 )}{( 2 x + 3 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} - 3 x ^{2} + 1}{( 2 x + 3 ) ^{4}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{4} - 3 x ^{2} + 1 ) ( 2 x + 3 ) ^{4} - \frac{d}{d x} ( ( 2 x + 3 ) ^{4} ) ( x ^{4} - 3 x ^{2} + 1 )}{( ( 2 x + 3 ) ^{4} ) ^{2}}

\frac{d}{d x} (x^{4} - 3 x^{2} + 1) = 4 x^{3} - 6 x

\frac{d}{d x} ((2 x + 3)^{4}) = 8 (2 x + 3)^{3}

= \frac{( 4 x ^{3} - 6 x ) ( 2 x + 3 ) ^{4} - 8 ( 2 x + 3 ) ^{3} ( x ^{4} - 3 x ^{2} + 1 )}{( ( 2 x + 3 ) ^{4} ) ^{2}}

Simplify

\frac{( 4 x ^{3} - 6 x ) ( 2 x + 3 ) ^{4} - 8 ( 2 x + 3 ) ^{3} ( x ^{4} - 3 x ^{2} + 1 )}{( ( 2 x + 3 ) ^{4} ) ^{2}} : \frac{2 ( 6 x ^{3} + 6 x ^{2} - 9 x - 4 )}{( 2 x + 3 ) ^{5}}

= \frac{2 ( 6 x ^{3} + 6 x ^{2} - 9 x - 4 )}{( 2 x + 3 ) ^{5}}

\frac{d}{d x} (\frac{2 x}{2})

4 x^{2} + 16 y^{2} - 51

\frac{\partial}{\partial x} (ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

x \to - 2 lim (x^{2} + 3 x - 1)

\int 2 \cdot sin (x) \cdot cos (x) d x

What is the derivative of (x^4-3x^2+1/((2x+3)^4)) ?
The derivative of (x^4-3x^2+1/((2x+3)^4)) is (2(6x^3+6x^2-9x-4))/((2x+3)^5)
What is the first derivative of (x^4-3x^2+1/((2x+3)^4)) ?
The first derivative of (x^4-3x^2+1/((2x+3)^4)) is (2(6x^3+6x^2-9x-4))/((2x+3)^5)