What is the derivative of f(x)=((x^2+8))/(7x-1) ?

The derivative of f(x)=((x^2+8))/(7x-1) is (7x^2-2x-56)/((7x-1)^2)

What is the first derivative of f(x)=((x^2+8))/(7x-1) ?

The first derivative of f(x)=((x^2+8))/(7x-1) is (7x^2-2x-56)/((7x-1)^2)

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derivative of

f (x) = \frac{( x ^{2} + 8 )}{7 x - 1}

\frac{7 x ^{2} - 2 x - 56}{( 7 x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 8}{7 x - 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 8 ) ( 7 x - 1 ) - \frac{d}{d x} ( 7 x - 1 ) ( x ^{2} + 8 )}{( 7 x - 1 ) ^{2}}

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

\frac{d}{d x} (7 x - 1) = 7

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

Expand

2 x (7 x - 1) - 7 (x^{2} + 8) : 7 x^{2} - 2 x - 56

= \frac{7 x ^{2} - 2 x - 56}{( 7 x - 1 ) ^{2}}

\int \frac{x}{6 x ^{2} + 1} d x

(2 x + 2 y)^{3} = 8 x^{3} + 8 y^{3}

\int (\frac{tan ( x )}{sin ( x ) cos ( x )}) d x

f (x) \frac{3}{x}

f (x) = 2 x^{2} + 3, x, [- 1, 2]

What is the derivative of f(x)=((x^2+8))/(7x-1) ?
The derivative of f(x)=((x^2+8))/(7x-1) is (7x^2-2x-56)/((7x-1)^2)
What is the first derivative of f(x)=((x^2+8))/(7x-1) ?
The first derivative of f(x)=((x^2+8))/(7x-1) is (7x^2-2x-56)/((7x-1)^2)