Solve 8/x^2-16-7/x^2-x-12

$$\frac { 8 } { x ^ { 2 } - 16 } - \frac { 7 } { x ^ { 2 } - x - 12 }$$

Evaluate

$$\frac{1}{\left(x+3\right)\left(x+4\right)}$$

Short Solution Steps

Factor x^{2}-16. Factor x^{2}-x-12.

$$\frac{8}{\left(x-4\right)\left(x+4\right)}-\frac{7}{\left(x-4\right)\left(x+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+4\right) and \left(x-4\right)\left(x+3\right) is \left(x-4\right)\left(x+3\right)\left(x+4\right). Multiply \frac{8}{\left(x-4\right)\left(x+4\right)} times \frac{x+3}{x+3}. Multiply \frac{7}{\left(x-4\right)\left(x+3\right)} times \frac{x+4}{x+4}.

$$\frac{8\left(x+3\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}-\frac{7\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Since \frac{8\left(x+3\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)} and \frac{7\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8\left(x+3\right)-7\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Do the multiplications in 8\left(x+3\right)-7\left(x+4\right).

$$\frac{8x+24-7x-28}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Combine like terms in 8x+24-7x-28.

$$\frac{x-4}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Cancel out x-4 in both numerator and denominator.

$$\frac{1}{\left(x+3\right)\left(x+4\right)}$$

Expand \left(x+3\right)\left(x+4\right).

$$\frac{1}{x^{2}+7x+12}$$

Differentiate w.r.t. x

$$\frac{-2x-7}{\left(\left(x+3\right)\left(x+4\right)\right)^{2}}$$

Steps Using Chain Rule

Factor x^{2}-16. Factor x^{2}-x-12.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8}{\left(x-4\right)\left(x+4\right)}-\frac{7}{\left(x-4\right)\left(x+3\right)})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+4\right) and \left(x-4\right)\left(x+3\right) is \left(x-4\right)\left(x+3\right)\left(x+4\right). Multiply \frac{8}{\left(x-4\right)\left(x+4\right)} times \frac{x+3}{x+3}. Multiply \frac{7}{\left(x-4\right)\left(x+3\right)} times \frac{x+4}{x+4}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8\left(x+3\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}-\frac{7\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)})$$

Since \frac{8\left(x+3\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)} and \frac{7\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8\left(x+3\right)-7\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)})$$

Do the multiplications in 8\left(x+3\right)-7\left(x+4\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x+24-7x-28}{\left(x-4\right)\left(x+3\right)\left(x+4\right)})$$

Combine like terms in 8x+24-7x-28.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-4}{\left(x-4\right)\left(x+3\right)\left(x+4\right)})$$

Cancel out x-4 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x+3\right)\left(x+4\right)})$$

Use the distributive property to multiply x+3 by x+4 and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+7x+12})$$

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).

$$-\left(x^{2}+7x^{1}+12\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+7x^{1}+12)$$

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}.

$$-\left(x^{2}+7x^{1}+12\right)^{-2}\left(2x^{2-1}+7x^{1-1}\right)$$

Simplify.

$$\left(x^{2}+7x^{1}+12\right)^{-2}\left(-2x^{1}-7x^{0}\right)$$

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

$$\left(x^{2}+7x+12\right)^{-2}\left(-2x-7x^{0}\right)$$

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

$$\left(x^{2}+7x+12\right)^{-2}\left(-2x-7\right)$$