RaySix

Post

at November 1st 2023, 4:29:24 pm.

Find the derivative of the function:

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

To find the derivative of the given function, we will use the quotient rule. The quotient rule states that for two functions $u(x)$ and $v(x)$ ,

\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}

Let's first find the derivatives of the numerator and denominator separately.

Numerator ( $u(x)$ ): Using the power rule and the sum/difference rule, we have:

u(x) = 3x^2 - 5x + 2

u'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(2)

u'(x) = 6x - 5

Denominator ( $v(x)$ ): Using the power rule and the sum/difference rule, we have:

v(x) = 4x^3 + 2x - 1

v'(x) = \frac{d}{dx}(4x^3) + \frac{d}{dx}(2x) - \frac{d}{dx}(1)

v'(x) = 12x^2 + 2

Using the quotient rule, we can now find the derivative of the function:

f'(x) = \frac{(u'(x)v(x) - u(x)v'(x))}{(v(x))^2}

f'(x) = \frac{(6x - 5)(4x^3 + 2x - 1) - (3x^2 - 5x + 2)(12x^2 + 2)}{(4x^3 + 2x - 1)^2}

Expanding and simplifying the numerator:

f'(x) = \frac{24x^4 + 12x^2 - 6x - 20x^3 - 10x + 5 - 36x^4 - 6x^2 + 60x^2 + 10x - 24x^2 - 4}{(4x^3 + 2x - 1)^2}

f'(x) = \frac{-12x^4 + 36x^3 + 32x^2 - 26x + 1}{(4x^3 + 2x - 1)^2}

Simplifying the expression, we get the final answer:

f'(x) = \frac{-12x^4 + 36x^3 + 32x^2 - 26x + 1}{(4x^3 + 2x - 1)^2}

Therefore, the derivative of the given function is $\frac{-12x^4 + 36x^3 + 32x^2 - 26x + 1}{(4x^3 + 2x - 1)^2}$ .