Solve 6/a^2-9+1/a+3

$$\frac { 6 } { a ^ { 2 } - 9 } + \frac { 1 } { a + 3 }$$

Evaluate

$$\frac{1}{a-3}$$

Short Solution Steps

Factor a^{2}-9.

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

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

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

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

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

Combine like terms in 6+a-3.

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

Cancel out a+3 in both numerator and denominator.

$$\frac{1}{a-3}$$

Differentiate w.r.t. a

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

Steps Using Chain Rule

Factor a^{2}-9.

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

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

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

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

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

Combine like terms in 6+a-3.

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

Cancel out a+3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{a-3})$$

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).

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

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}.

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

Simplify.

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

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

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

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

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