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

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

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

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

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derivative of

\frac{3}{x ^{4} + 7 x + 1}

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

Solution steps

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

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{1}{x ^{4} + 7 x + 1})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{4} + 7 x + 1) = 4 x^{3} + 7

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

Simplify

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

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

y^{^{'}} = 3 x y

\frac{7 + 2 x}{1 - 6 x}

\frac{x ^{2} - 6 x - 1}{( x - 3 ) ^{2}}

x \to 1 lim ([2 f (x) + g (x)])

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

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