What is the derivative of f(x)= 1/2 (7x^2+x)^{-3} ?

The derivative of f(x)= 1/2 (7x^2+x)^{-3} is -(3(14x+1))/(2(7x^2+x)^4)

What is the first derivative of f(x)= 1/2 (7x^2+x)^{-3} ?

The first derivative of f(x)= 1/2 (7x^2+x)^{-3} is -(3(14x+1))/(2(7x^2+x)^4)

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derivative of

f (x) = \frac{1}{2} (7 x^{2} + x)^{- 3}

- \frac{3 ( 14 x + 1 )}{2 ( 7 x ^{2} + x ) ^{4}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

= \frac{1}{2} (- \frac{3}{( 7 x ^{2} + x ) ^{4}} (14 x + 1))

Simplify

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

= - \frac{3 ( 14 x + 1 )}{2 ( 7 x ^{2} + x ) ^{4}}

\int_{0}^{1} \int_{4 y}^{4} e^{x^{2}} d x d y

\int \frac{x ^{2}}{( x + 3 ) ^{3}} d x

\frac{d v}{d t} + \frac{( v ^{^{'}} \cdot v )}{t} = v^{^{'}} \cdot (\frac{t}{2} - \frac{1}{2})

\frac{d}{d x} (2 x ln (x))

What is the derivative of f(x)= 1/2 (7x^2+x)^{-3} ?
The derivative of f(x)= 1/2 (7x^2+x)^{-3} is -(3(14x+1))/(2(7x^2+x)^4)
What is the first derivative of f(x)= 1/2 (7x^2+x)^{-3} ?
The first derivative of f(x)= 1/2 (7x^2+x)^{-3} is -(3(14x+1))/(2(7x^2+x)^4)