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

The derivative of 1/((-x^2+x-2^5)) is -(5(-2x+1))/((-x^2+x-2)^6)

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

The first derivative of 1/((-x^2+x-2^5)) is -(5(-2x+1))/((-x^2+x-2)^6)

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

- \frac{5 ( - 2 x + 1 )}{( - x ^{2} + x - 2 ) ^{6}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= - \frac{5}{( - x ^{2} + x - 2 ) ^{6}} (- 2 x + 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{5 ( - 2 x + 1 )}{( - x ^{2} + x - 2 ) ^{6}}

x^{2} y^{^{''}} + 5 x y^{^{'}} + 4 y = 0

\frac{d y}{d x} = \frac{1}{x ^{2}} - \frac{2 y}{x}

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

\frac{d}{d θ} (2 cos (θ) + 2 sin (θ))

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

What is the derivative of 1/((-x^2+x-2^5)) ?
The derivative of 1/((-x^2+x-2^5)) is -(5(-2x+1))/((-x^2+x-2)^6)
What is the first derivative of 1/((-x^2+x-2^5)) ?
The first derivative of 1/((-x^2+x-2^5)) is -(5(-2x+1))/((-x^2+x-2)^6)