Solve h(x)=-x/3x^2-9

$$h ( x ) = \frac { - x } { 3 x ^ { 2 } - 9 }$$

Differentiate w.r.t. x

$$\frac{x^{2}+3}{3\left(x^{2}-3\right)^{2}}$$

Steps Using Derivative Rule for Quotient

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

$$\frac{\left(3x^{2}-9\right)\frac{\mathrm{d}}{\mathrm{d}x}(-x^{1})-\left(-x^{1}\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}-9)\right)}{\left(3x^{2}-9\right)^{2}}$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$\frac{\left(3x^{2}-9\right)\left(-1\right)x^{1-1}-\left(-x^{1}\times 2\times 3x^{2-1}\right)}{\left(3x^{2}-9\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(3x^{2}-9\right)\left(-1\right)x^{0}-\left(-x^{1}\times 6x^{1}\right)}{\left(3x^{2}-9\right)^{2}}$$

Expand using distributive property.

$$\frac{3x^{2}\left(-1\right)x^{0}-9\left(-1\right)x^{0}-\left(-x^{1}\times 6x^{1}\right)}{\left(3x^{2}-9\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{3\left(-1\right)x^{2}-9\left(-1\right)x^{0}-\left(-6x^{1+1}\right)}{\left(3x^{2}-9\right)^{2}}$$

Do the arithmetic.

$$\frac{-3x^{2}+9x^{0}-\left(-6x^{2}\right)}{\left(3x^{2}-9\right)^{2}}$$

Combine like terms.

$$\frac{\left(-3-\left(-6\right)\right)x^{2}+9x^{0}}{\left(3x^{2}-9\right)^{2}}$$

Subtract -6 from -3.

$$\frac{3x^{2}+9x^{0}}{\left(3x^{2}-9\right)^{2}}$$

Factor out 3.

$$\frac{3\left(x^{2}+3x^{0}\right)}{\left(3x^{2}-9\right)^{2}}$$

For any term t except 0, t^{0}=1.

$$\frac{3\left(x^{2}+3\times 1\right)}{\left(3x^{2}-9\right)^{2}}$$

For any term t, t\times 1=t and 1t=t.

$$\frac{3\left(x^{2}+3\right)}{\left(3x^{2}-9\right)^{2}}$$

Evaluate

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