Quotient Rule

Step 1: Simplify the numerator and denominator.

Both the numerator and denominator are already simplified.

$\frac{x^3-5x+8}{x^2-15}$

Step 2: Apply the quotient rule.

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^2}$

$\frac{d}{dx}\left[\frac{x^3-5x+8}{x^2-15}\right]$

$\frac{\left[\left(x^2-15\right)\frac{d}{dx}\left(x^3-5x+8\right)\right]-\left[\left(x^3-5x+8\right)\frac{d}{dx}\left(x^2-15\right)\right]}{{\left(x^2-15\right)}^2}$

Step 3: Calculate the two derivatives.

$\frac{d}{dx}\left(x^3-5x+8\right)$ Original

$\frac{d}{dx}{x}^3-\frac{d}{dx}5x+\frac{d}{dx}8$    Sum/Diff Rule

$\frac{d}{dx}{x}^3-5\frac{d}{dx}x+\frac{d}{dx}8$    Constant Multiple

3x² - 5(1) + 0 Power & Constant

$3x^2-5$

______________________________

$\frac{d}{dx}\left(x^2-15\right)$ Original

$\frac{d}{dx}{x}^2-\frac{d}{dx}15$   Difference Rule

2x¹ - 0 Power & Constant

$2x$

Step 4: Substitute the derivatives & simplify.

$\frac{\left[\left(x^2-15\right)\left(3x^2-5\right)\right]-\left[\left(x^3-5x+8\right)\left(2x\right)\right]}{{\left(x^2-15\right)}^2}$

$\frac{\left(3x^4-5x^2-45x^2+75\right)-\left(2x^4-10x^2+16x\right)}{{\left(x^2-15\right)}^2}$

$\frac{\left(3x^4-50x^2+75\right)+\left(-2x^4+10x^2-16x\right)}{{\left(x^2-15\right)}^2}$

$\frac{x^4-40x^2-16x+75}{{\left(x^2-15\right)}^2}$

Step 1: Simplify the numerator and denominator.

Both the numerator and denominator are already simplified.

$\frac{3e^x}{\sqrt{x}}$

Step 2: Apply the quotient rule.

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^2}$

$\frac{d}{dx}\left[\frac{3e^x}{\sqrt{x}}\right]$

$\frac{\left[\left(\sqrt{x}\right)\frac{d}{dx}\left(3e^x\right)\right]-\left[\left(3e^x\right)\frac{d}{dx}\left(\sqrt{x}\right)\right]}{{\left(\sqrt{x}\right)}^2}$

Step 3: Calculate the two derivatives.

$\frac{d}{dx}\left(3e^x\right)$      Original

$3\frac{d}{dx}{e}^x$    Constant Mult.

3e^x        Natural Exp.

$3e^x$

______________________________

$\frac{d}{dx}\left(\sqrt{x}\right)$    Original

$\frac{1}{2}{x}^{-\frac{1}{2}}$       Power

$\frac{1}{2\sqrt{x}}$

Step 4: Substitute the derivatives & simplify.

$\frac{\left[\left(\sqrt{x}\right)3e^x\right]-\left[\left(3e^x\right)\frac{1}{2\sqrt{x}}\right] }{{\left(\sqrt{x}\right)}^2}$

$\frac{3e^x\left(\sqrt{x}-\frac{1}{2\sqrt{x}}\right)}{x}$

Step 1: Simplify the numerator and denominator.

Both the numerator and denominator are already simplified.

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

Step 2: Apply the quotient rule.

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^2}$

$\frac{d}{dx}\left[\frac{2x^3}{1-3x^2}\right]$

$\frac{\left[\left(1-3x^2\right)\frac{d}{dx}\left(2x^3\right)\right]-\left[\left(2x^3\right)\frac{d}{dx}\left(1-3x^2\right)\right]}{{\left(1-3x^2\right)}^2}$

Step 3: Calculate the two derivatives.

$\frac{d}{dx}\left(2x^3\right)$    Original

$2\frac{d}{dx}{x}^3$    Constant Multiple

2(3x²)      Power

$6x^2$

________________________

$\frac{d}{dx}\left(1-3x^2\right)$ Original

$\frac{d}{dx}1-\frac{d}{dx}3x^2$ Difference Rule

$\frac{d}{dx}1-3\frac{d}{dx}{x}^2$ Constant Multiple

0 - 3(2x) Constant/Power

$-6x$

Step 4: Substitute the derivatives & simplify.

$\frac{\left[\left(1-3x^2\right)6x^2\right]-\left[\left(2x^3\right)\left(-6x\right)\right]}{{\left(1-3x^2\right)}^2}$

$\frac{6x^2-18x^4+12x^4}{{\left(1-3x^2\right)}^2}$

$\frac{6x^2-6x^4}{{\left(1-3x^2\right)}^2}$