derivative of (2x^2-3x+1/x) — step-by-step solution

To simplify differentiation, we rewrite

\frac{1}{x}

x^{-1}

Let's differentiate:

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

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

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

We differentiate using:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

(where f =

x

)

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

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

2 \cdot \frac{d}{d x} (x^{2})

\frac{d}{d x} (- 3 x)

-3

■The derivative of x^n is n*x^(n-1):

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

2 x

\frac{d}{d x} (x^{-1})

-1 \cdot x^{-2}

2 \cdot 2 x - 3 - 1 \cdot x^{-2}

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 2

4

■Multiplying by 1 changes nothing:

1 \cdot x^{-2}

x^{-2}

4 x - 3 - x^{-2}

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

x^{-2}

\frac{1}{x^{2}}

4 x - 3 - \frac{1}{x^{2}}

▶Derivative:

4 x - \frac{1}{x^{2}} - 3