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

The derivative of f(x)=(7x^2+7x+7)/(sqrt(x)) is (21x^2+7x-7)/(2xsqrt(x))

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

The first derivative of f(x)=(7x^2+7x+7)/(sqrt(x)) is (21x^2+7x-7)/(2xsqrt(x))

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derivative of

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

\frac{21 x ^{2} + 7 x - 7}{2 x x}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (7 x^{2} + 7 x + 7) = 14 x + 7

\frac{d}{d x} (x) = \frac{1}{2 x}

= \frac{( 14 x + 7 ) x - \frac{1}{2 x} ( 7 x ^{2} + 7 x + 7 )}{( x ) ^{2}}

Simplify

\frac{( 14 x + 7 ) x - \frac{1}{2 x} ( 7 x ^{2} + 7 x + 7 )}{( x ) ^{2}} : \frac{21 x ^{2} + 7 x - 7}{2 x x}

= \frac{21 x ^{2} + 7 x - 7}{2 x x}

y = (x^{(3)} - 4 x)^{(12)} (2)

\frac{\partial}{\partial y} (y + cos^{2} (x))

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

n = 1 \sum \infty tan (3 n)

g (x) = \frac{x ^{2} - 64}{x + 8}

What is the derivative of f(x)=(7x^2+7x+7)/(sqrt(x)) ?
The derivative of f(x)=(7x^2+7x+7)/(sqrt(x)) is (21x^2+7x-7)/(2xsqrt(x))
What is the first derivative of f(x)=(7x^2+7x+7)/(sqrt(x)) ?
The first derivative of f(x)=(7x^2+7x+7)/(sqrt(x)) is (21x^2+7x-7)/(2xsqrt(x))