What is the derivative of 7(10x^5-2x+6)^{-3} ?

The derivative of 7(10x^5-2x+6)^{-3} is -(21(25x^4-1))/(8(5x^5-x+3)^4)

What is the first derivative of 7(10x^5-2x+6)^{-3} ?

The first derivative of 7(10x^5-2x+6)^{-3} is -(21(25x^4-1))/(8(5x^5-x+3)^4)

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derivative of

7 (10 x^{5} - 2 x + 6)^{- 3}

- \frac{21 ( 25 x ^{4} - 1 )}{8 ( 5 x ^{5} - x + 3 ) ^{4}}

Solution steps

\frac{d}{d x} (7 (10 x^{5} - 2 x + 6)^{- 3})

Take the constant out:

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

= 7 \frac{d}{d x} ((10 x^{5} - 2 x + 6)^{- 3})

Apply the chain rule:

- \frac{3}{( 10 x ^{5} - 2 x + 6 ) ^{4}} \frac{d}{d x} (10 x^{5} - 2 x + 6)

= - \frac{3}{( 10 x ^{5} - 2 x + 6 ) ^{4}} \frac{d}{d x} (10 x^{5} - 2 x + 6)

\frac{d}{d x} (10 x^{5} - 2 x + 6) = 50 x^{4} - 2

= 7 (- \frac{3}{( 10 x ^{5} - 2 x + 6 ) ^{4}} (50 x^{4} - 2))

Simplify

7 (- \frac{3}{( 10 x ^{5} - 2 x + 6 ) ^{4}} (50 x^{4} - 2)) : - \frac{21 ( 25 x ^{4} - 1 )}{8 ( 5 x ^{5} - x + 3 ) ^{4}}

= - \frac{21 ( 25 x ^{4} - 1 )}{8 ( 5 x ^{5} - x + 3 ) ^{4}}

\frac{\partial}{\partial s} (s - t)

\frac{d}{d y} (\frac{x y}{x - y})

\int \frac{4 cosh ( 5 x )}{sinh ^{3} ( 5 x )} d x

f (x) = 3 x^{2} - 4 x + 5

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

What is the derivative of 7(10x^5-2x+6)^{-3} ?
The derivative of 7(10x^5-2x+6)^{-3} is -(21(25x^4-1))/(8(5x^5-x+3)^4)
What is the first derivative of 7(10x^5-2x+6)^{-3} ?
The first derivative of 7(10x^5-2x+6)^{-3} is -(21(25x^4-1))/(8(5x^5-x+3)^4)