Solve (1/x-3-6/x^2-9)

$$( \frac { 1 } { x - 3 } - \frac { 6 } { x ^ { 2 } - 9 } )$$

Evaluate

$$\frac{1}{x+3}$$

Short Solution Steps

Factor x^{2}-9.

$$\frac{1}{x-3}-\frac{6}{\left(x-3\right)\left(x+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-3 and \left(x-3\right)\left(x+3\right) is \left(x-3\right)\left(x+3\right). Multiply \frac{1}{x-3} times \frac{x+3}{x+3}.

$$\frac{x+3}{\left(x-3\right)\left(x+3\right)}-\frac{6}{\left(x-3\right)\left(x+3\right)}$$

Since \frac{x+3}{\left(x-3\right)\left(x+3\right)} and \frac{6}{\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+3-6}{\left(x-3\right)\left(x+3\right)}$$

Combine like terms in x+3-6.

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

Cancel out x-3 in both numerator and denominator.

$$\frac{1}{x+3}$$

Differentiate w.r.t. x

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

Steps Using Chain Rule

Factor x^{2}-9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x-3}-\frac{6}{\left(x-3\right)\left(x+3\right)})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-3 and \left(x-3\right)\left(x+3\right) is \left(x-3\right)\left(x+3\right). Multiply \frac{1}{x-3} times \frac{x+3}{x+3}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+3}{\left(x-3\right)\left(x+3\right)}-\frac{6}{\left(x-3\right)\left(x+3\right)})$$

Since \frac{x+3}{\left(x-3\right)\left(x+3\right)} and \frac{6}{\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+3-6}{\left(x-3\right)\left(x+3\right)})$$

Combine like terms in x+3-6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-3}{\left(x-3\right)\left(x+3\right)})$$

Cancel out x-3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x+3})$$

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

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}.

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

Simplify.

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

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

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

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

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