Solve 1/x+6-1/x+7

$$\frac { 1 } { x + 6 } - \frac { 1 } { x + 7 }$$

Evaluate

$$\frac{1}{\left(x+6\right)\left(x+7\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+6 and x+7 is \left(x+6\right)\left(x+7\right). Multiply \frac{1}{x+6} times \frac{x+7}{x+7}. Multiply \frac{1}{x+7} times \frac{x+6}{x+6}.

$$\frac{x+7}{\left(x+6\right)\left(x+7\right)}-\frac{x+6}{\left(x+6\right)\left(x+7\right)}$$

Since \frac{x+7}{\left(x+6\right)\left(x+7\right)} and \frac{x+6}{\left(x+6\right)\left(x+7\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+7-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)}$$

Do the multiplications in x+7-\left(x+6\right).

$$\frac{x+7-x-6}{\left(x+6\right)\left(x+7\right)}$$

Combine like terms in x+7-x-6.

$$\frac{1}{\left(x+6\right)\left(x+7\right)}$$

Expand \left(x+6\right)\left(x+7\right).

$$\frac{1}{x^{2}+13x+42}$$

Differentiate w.r.t. x

$$\frac{-2x-13}{\left(\left(x+6\right)\left(x+7\right)\right)^{2}}$$

Steps Using Chain Rule

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+6 and x+7 is \left(x+6\right)\left(x+7\right). Multiply \frac{1}{x+6} times \frac{x+7}{x+7}. Multiply \frac{1}{x+7} times \frac{x+6}{x+6}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+7}{\left(x+6\right)\left(x+7\right)}-\frac{x+6}{\left(x+6\right)\left(x+7\right)})$$

Since \frac{x+7}{\left(x+6\right)\left(x+7\right)} and \frac{x+6}{\left(x+6\right)\left(x+7\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+7-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)})$$

Do the multiplications in x+7-\left(x+6\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+7-x-6}{\left(x+6\right)\left(x+7\right)})$$

Combine like terms in x+7-x-6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x+6\right)\left(x+7\right)})$$

Apply the distributive property by multiplying each term of x+6 by each term of x+7.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+7x+6x+42})$$

Combine 7x and 6x to get 13x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+13x+42})$$

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).

$$-\left(x^{2}+13x^{1}+42\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+13x^{1}+42)$$

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}.

$$-\left(x^{2}+13x^{1}+42\right)^{-2}\left(2x^{2-1}+13x^{1-1}\right)$$

Simplify.

$$\left(x^{2}+13x^{1}+42\right)^{-2}\left(-2x^{1}-13x^{0}\right)$$

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

$$\left(x^{2}+13x+42\right)^{-2}\left(-2x-13x^{0}\right)$$

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

$$\left(x^{2}+13x+42\right)^{-2}\left(-2x-13\right)$$