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

The derivative of h(x)=(x^2+4x+1)/(3x+5) is (3x^2+10x+17)/((3x+5)^2)

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

The first derivative of h(x)=(x^2+4x+1)/(3x+5) is (3x^2+10x+17)/((3x+5)^2)

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derivative of

h (x) = \frac{x ^{2} + 4 x + 1}{3 x + 5}

\frac{3 x ^{2} + 10 x + 17}{( 3 x + 5 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (3 x + 5) = 3

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

Expand

(2 x + 4) (3 x + 5) - 3 (x^{2} + 4 x + 1) : 3 x^{2} + 10 x + 17

= \frac{3 x ^{2} + 10 x + 17}{( 3 x + 5 ) ^{2}}

\frac{d}{d x} (5 x^{2 x})

\int 2 x \cdot e^{2 x} d x

\int (1 + \frac{1}{u})^{- 3} (\frac{1}{u ^{2}}) d u

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

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{36}

What is the derivative of h(x)=(x^2+4x+1)/(3x+5) ?
The derivative of h(x)=(x^2+4x+1)/(3x+5) is (3x^2+10x+17)/((3x+5)^2)
What is the first derivative of h(x)=(x^2+4x+1)/(3x+5) ?
The first derivative of h(x)=(x^2+4x+1)/(3x+5) is (3x^2+10x+17)/((3x+5)^2)