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

The derivative of ((-2x-3)^7)/((4x+2)^4) is -((-2x-3)^6(6x-5))/(8(2x+1)^5)

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

The first derivative of ((-2x-3)^7)/((4x+2)^4) is -((-2x-3)^6(6x-5))/(8(2x+1)^5)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} ((- 2 x - 3)^{7}) = - 14 (- 2 x - 3)^{6}

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

= \frac{( - 14 ( - 2 x - 3 ) ^{6} ) ( 4 x + 2 ) ^{4} - 16 ( 4 x + 2 ) ^{3} ( - 2 x - 3 ) ^{7}}{( ( 4 x + 2 ) ^{4} ) ^{2}}

Simplify

\frac{( - 14 ( - 2 x - 3 ) ^{6} ) ( 4 x + 2 ) ^{4} - 16 ( 4 x + 2 ) ^{3} ( - 2 x - 3 ) ^{7}}{( ( 4 x + 2 ) ^{4} ) ^{2}} : - \frac{( - 2 x - 3 ) ^{6} ( 6 x - 5 )}{8 ( 2 x + 1 ) ^{5}}

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

t \to \infty lim (x)

\int_{0}^{9} π e^{- \frac{2 x}{9}} d x

\int x^{3} x^{2} + 11 d x

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{x}

\int_{0}^{5} \frac{1}{25 - x ^{2}} d x

What is the derivative of ((-2x-3)^7)/((4x+2)^4) ?
The derivative of ((-2x-3)^7)/((4x+2)^4) is -((-2x-3)^6(6x-5))/(8(2x+1)^5)
What is the first derivative of ((-2x-3)^7)/((4x+2)^4) ?
The first derivative of ((-2x-3)^7)/((4x+2)^4) is -((-2x-3)^6(6x-5))/(8(2x+1)^5)