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

The derivative of (4x^2-5x+1/(3x+7)) is (12x^2+56x-38)/((3x+7)^2)

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

The first derivative of (4x^2-5x+1/(3x+7)) is (12x^2+56x-38)/((3x+7)^2)

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\frac{d}{d x} (\frac{4 x ^{2} - 5 x + 1}{3 x + 7})

\frac{12 x ^{2} + 56 x - 38}{( 3 x + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{2} - 5 x + 1}{3 x + 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} - 5 x + 1 ) ( 3 x + 7 ) - \frac{d}{d x} ( 3 x + 7 ) ( 4 x ^{2} - 5 x + 1 )}{( 3 x + 7 ) ^{2}}

\frac{d}{d x} (4 x^{2} - 5 x + 1) = 8 x - 5

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

= \frac{( 8 x - 5 ) ( 3 x + 7 ) - 3 ( 4 x ^{2} - 5 x + 1 )}{( 3 x + 7 ) ^{2}}

Expand

(8 x - 5) (3 x + 7) - 3 (4 x^{2} - 5 x + 1) : 12 x^{2} + 56 x - 38

= \frac{12 x ^{2} + 56 x - 38}{( 3 x + 7 ) ^{2}}

\int_{1}^{4} (5 x^{2} + x) d x

csc (y) + sec^{2} (x) (\frac{d y}{d x}) = 0

n = 1 \sum \infty \frac{3}{n ( n + 1 )}

f (x) = x, g (x) = x, x = 0, 2

x y^{^{'}} = 8 y - 9 x, y (1) = 0, y (1) = 0

What is the derivative of (4x^2-5x+1/(3x+7)) ?
The derivative of (4x^2-5x+1/(3x+7)) is (12x^2+56x-38)/((3x+7)^2)
What is the first derivative of (4x^2-5x+1/(3x+7)) ?
The first derivative of (4x^2-5x+1/(3x+7)) is (12x^2+56x-38)/((3x+7)^2)