What is the derivative of ((4x-2^4)/(3x^2+7)) ?

The derivative of ((4x-2^4)/(3x^2+7)) is (16(4x-2)^3(3x^2+7)-6x(4x-2)^4)/((3x^2+7)^2)

What is the first derivative of ((4x-2^4)/(3x^2+7)) ?

The first derivative of ((4x-2^4)/(3x^2+7)) is (16(4x-2)^3(3x^2+7)-6x(4x-2)^4)/((3x^2+7)^2)

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\frac{d}{d x} (\frac{( 4 x - 2 ) ^{4}}{3 x ^{2} + 7})

\frac{16 ( 4 x - 2 ) ^{3} ( 3 x ^{2} + 7 ) - 6 x ( 4 x - 2 ) ^{4}}{( 3 x ^{2} + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( 4 x - 2 ) ^{4}}{3 x ^{2} + 7})

Apply the Quotient Rule:

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

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

\frac{d}{d x} ((4 x - 2)^{4}) = 16 (4 x - 2)^{3}

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

= \frac{16 ( 4 x - 2 ) ^{3} ( 3 x ^{2} + 7 ) - 6 x ( 4 x - 2 ) ^{4}}{( 3 x ^{2} + 7 ) ^{2}}

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

x y^{^{'}} = y + (x y)^{\frac{1}{2}}

\frac{d y}{d x} + y = 10

\int_{1}^{4} (x) d x

\frac{d ^{3}}{d x ^{3}} (e^{x} cos (x))

What is the derivative of ((4x-2^4)/(3x^2+7)) ?
The derivative of ((4x-2^4)/(3x^2+7)) is (16(4x-2)^3(3x^2+7)-6x(4x-2)^4)/((3x^2+7)^2)
What is the first derivative of ((4x-2^4)/(3x^2+7)) ?
The first derivative of ((4x-2^4)/(3x^2+7)) is (16(4x-2)^3(3x^2+7)-6x(4x-2)^4)/((3x^2+7)^2)