What is the derivative of-7(3x^4+2^{-4}) ?

The derivative of-7(3x^4+2^{-4}) is (336x^3)/((3x^4+2)^5)

What is the first derivative of-7(3x^4+2^{-4}) ?

The first derivative of-7(3x^4+2^{-4}) is (336x^3)/((3x^4+2)^5)

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

\frac{336 x ^{3}}{( 3 x ^{4} + 2 ) ^{5}}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{4} + 2) = 12 x^{3}

= - 7 (- \frac{4}{( 3 x ^{4} + 2 ) ^{5}} \cdot 12 x^{3})

Simplify

- 7 (- \frac{4}{( 3 x ^{4} + 2 ) ^{5}} \cdot 12 x^{3}) : \frac{336 x ^{3}}{( 3 x ^{4} + 2 ) ^{5}}

= \frac{336 x ^{3}}{( 3 x ^{4} + 2 ) ^{5}}

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

x \to 5 lim (\frac{x + 1}{x - 5})

(x^{2} - 8)^{4}, at x = 3

f (x) = 7 + 4 x^{2} - 2 x^{3}, at x = 1

\frac{d}{d x} (c e^{x} + c e^{- x})

What is the derivative of-7(3x^4+2^{-4}) ?
The derivative of-7(3x^4+2^{-4}) is (336x^3)/((3x^4+2)^5)
What is the first derivative of-7(3x^4+2^{-4}) ?
The first derivative of-7(3x^4+2^{-4}) is (336x^3)/((3x^4+2)^5)