What is the derivative of (2x^5+3x^4-1/(-2x^7)) ?

The derivative of (2x^5+3x^4-1/(-2x^7)) is -(-4x^5-9x^4+7)/(2x^8)

What is the first derivative of (2x^5+3x^4-1/(-2x^7)) ?

The first derivative of (2x^5+3x^4-1/(-2x^7)) is -(-4x^5-9x^4+7)/(2x^8)

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\frac{d}{d x} (\frac{2 x ^{5} + 3 x ^{4} - 1}{- 2 x ^{7}})

- \frac{- 4 x ^{5} - 9 x ^{4} + 7}{2 x ^{8}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{5} + 3 x ^{4} - 1}{- 2 x ^{7}})

Simplify

\frac{2 x ^{5} + 3 x ^{4} - 1}{- 2 x ^{7}} : - \frac{2 x ^{5} + 3 x ^{4} - 1}{2 x ^{7}}

= \frac{d}{d x} (- \frac{2 x ^{5} + 3 x ^{4} - 1}{2 x ^{7}})

Take the constant out:

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

= - \frac{1}{2} \frac{d}{d x} (\frac{2 x ^{5} + 3 x ^{4} - 1}{x ^{7}})

Apply the Quotient Rule:

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

= - \frac{1}{2} \cdot \frac{\frac{d}{d x} ( 2 x ^{5} + 3 x ^{4} - 1 ) x ^{7} - \frac{d}{d x} ( x ^{7} ) ( 2 x ^{5} + 3 x ^{4} - 1 )}{( x ^{7} ) ^{2}}

\frac{d}{d x} (2 x^{5} + 3 x^{4} - 1) = 10 x^{4} + 12 x^{3}

\frac{d}{d x} (x^{7}) = 7 x^{6}

= - \frac{1}{2} \cdot \frac{( 10 x ^{4} + 12 x ^{3} ) x ^{7} - 7 x ^{6} ( 2 x ^{5} + 3 x ^{4} - 1 )}{( x ^{7} ) ^{2}}

Simplify

- \frac{1}{2} \cdot \frac{( 10 x ^{4} + 12 x ^{3} ) x ^{7} - 7 x ^{6} ( 2 x ^{5} + 3 x ^{4} - 1 )}{( x ^{7} ) ^{2}} : - \frac{- 4 x ^{5} - 9 x ^{4} + 7}{2 x ^{8}}

= - \frac{- 4 x ^{5} - 9 x ^{4} + 7}{2 x ^{8}}

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

\frac{d}{d θ} (- cos (θ))

\int \frac{6}{2 x - 5} d x

n = 0 \sum \infty \frac{0 ^{n}}{3 ^{n}}

\frac{\partial}{\partial z} (z arctan (\frac{y}{x}))

What is the derivative of (2x^5+3x^4-1/(-2x^7)) ?
The derivative of (2x^5+3x^4-1/(-2x^7)) is -(-4x^5-9x^4+7)/(2x^8)
What is the first derivative of (2x^5+3x^4-1/(-2x^7)) ?
The first derivative of (2x^5+3x^4-1/(-2x^7)) is -(-4x^5-9x^4+7)/(2x^8)