Solve 2x/x^2-9+1/3-x

$$\frac { 2 x } { x ^ { 2 } - 9 } + \frac { 1 } { 3 - x }$$

Evaluate

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

Short Solution Steps

Factor x^{2}-9.

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

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

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

Since \frac{2x}{\left(x-3\right)\left(x+3\right)} and \frac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2x-\left(x+3\right).

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

Combine like terms in 2x-x-3.

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

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.

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

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

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

Since \frac{2x}{\left(x-3\right)\left(x+3\right)} and \frac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2x-\left(x+3\right).

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

Combine like terms in 2x-x-3.

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

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.

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