Solve 1/x+4-1/x+5

$$\frac { 1 } { x + 4 } - \frac { 1 } { x + 5 }$$

Evaluate

$$\frac{1}{\left(x+4\right)\left(x+5\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+4 and x+5 is \left(x+4\right)\left(x+5\right). Multiply \frac{1}{x+4} times \frac{x+5}{x+5}. Multiply \frac{1}{x+5} times \frac{x+4}{x+4}.

$$\frac{x+5}{\left(x+4\right)\left(x+5\right)}-\frac{x+4}{\left(x+4\right)\left(x+5\right)}$$

Since \frac{x+5}{\left(x+4\right)\left(x+5\right)} and \frac{x+4}{\left(x+4\right)\left(x+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+5-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}$$

Do the multiplications in x+5-\left(x+4\right).

$$\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}$$

Combine like terms in x+5-x-4.

$$\frac{1}{\left(x+4\right)\left(x+5\right)}$$

Expand \left(x+4\right)\left(x+5\right).

$$\frac{1}{x^{2}+9x+20}$$

Differentiate w.r.t. x

$$\frac{-2x-9}{\left(\left(x+4\right)\left(x+5\right)\right)^{2}}$$

Steps Using Chain Rule

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+4 and x+5 is \left(x+4\right)\left(x+5\right). Multiply \frac{1}{x+4} times \frac{x+5}{x+5}. Multiply \frac{1}{x+5} times \frac{x+4}{x+4}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+5}{\left(x+4\right)\left(x+5\right)}-\frac{x+4}{\left(x+4\right)\left(x+5\right)})$$

Since \frac{x+5}{\left(x+4\right)\left(x+5\right)} and \frac{x+4}{\left(x+4\right)\left(x+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+5-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)})$$

Do the multiplications in x+5-\left(x+4\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)})$$

Combine like terms in x+5-x-4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x+4\right)\left(x+5\right)})$$

Apply the distributive property by multiplying each term of x+4 by each term of x+5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+5x+4x+20})$$

Combine 5x and 4x to get 9x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+9x+20})$$

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).

$$-\left(x^{2}+9x^{1}+20\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+9x^{1}+20)$$

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}.

$$-\left(x^{2}+9x^{1}+20\right)^{-2}\left(2x^{2-1}+9x^{1-1}\right)$$

Simplify.

$$\left(x^{2}+9x^{1}+20\right)^{-2}\left(-2x^{1}-9x^{0}\right)$$

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

$$\left(x^{2}+9x+20\right)^{-2}\left(-2x-9x^{0}\right)$$

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

$$\left(x^{2}+9x+20\right)^{-2}\left(-2x-9\right)$$