1. **State the problem:** Find the derivative of the function $$y = \frac{5x^4 + 3x - 5 + 2}{x^3 + \pi x}$$. 2. **Recall the formula:** For a function $$y = \frac{u(x)}{v(x)}$$, the derivative is given by the quotient rule: $$ \frac{dy}{dx} = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2} $$ where $$u(x) = 5x^4 + 3x - 3$$ (since $$-5 + 2 = -3$$) and $$v(x) = x^3 + \pi x$$. 3. **Find the derivatives of numerator and denominator:** - $$u'(x) = \frac{d}{dx}(5x^4 + 3x - 3) = 20x^3 + 3$$ - $$v'(x) = \frac{d}{dx}(x^3 + \pi x) = 3x^2 + \pi$$ 4. **Apply the quotient rule:** $$ \frac{dy}{dx} = \frac{(x^3 + \pi x)(20x^3 + 3) - (5x^4 + 3x - 3)(3x^2 + \pi)}{(x^3 + \pi x)^2} $$ 5. **Expand the numerator:** - First term: $$ (x^3)(20x^3) = 20x^6 $$ $$ (x^3)(3) = 3x^3 $$ $$ (\pi x)(20x^3) = 20\pi x^4 $$ $$ (\pi x)(3) = 3\pi x $$ Sum first term: $$20x^6 + 3x^3 + 20\pi x^4 + 3\pi x$$ - Second term: $$ (5x^4)(3x^2) = 15x^6 $$ $$ (5x^4)(\pi) = 5\pi x^4 $$ $$ (3x)(3x^2) = 9x^3 $$ $$ (3x)(\pi) = 3\pi x $$ $$ (-3)(3x^2) = -9x^2 $$ $$ (-3)(\pi) = -3\pi $$ Sum second term: $$15x^6 + 5\pi x^4 + 9x^3 + 3\pi x - 9x^2 - 3\pi$$ 6. **Subtract second term from first term:** $$ (20x^6 + 3x^3 + 20\pi x^4 + 3\pi x) - (15x^6 + 5\pi x^4 + 9x^3 + 3\pi x - 9x^2 - 3\pi) $$ $$ = (20x^6 - 15x^6) + (3x^3 - 9x^3) + (20\pi x^4 - 5\pi x^4) + (3\pi x - 3\pi x) + 9x^2 + 3\pi $$ $$ = 5x^6 - 6x^3 + 15\pi x^4 + 0 + 9x^2 + 3\pi $$ 7. **Write the final derivative:** $$ \frac{dy}{dx} = \frac{5x^6 - 6x^3 + 15\pi x^4 + 9x^2 + 3\pi}{(x^3 + \pi x)^2} $$ This is the derivative of the given function.