Solve 1/x-4-1/x-3

$$\frac { 1 } { x - 4 } - \frac { 1 } { x - 3 }$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-4 and x-3 is \left(x-4\right)\left(x-3\right). Multiply \frac{1}{x-4} times \frac{x-3}{x-3}. Multiply \frac{1}{x-3} times \frac{x-4}{x-4}.

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

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

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

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

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

Combine like terms in x-3-x+4.

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

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

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

Differentiate w.r.t. x

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

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-3 is \left(x-4\right)\left(x-3\right). Multiply \frac{1}{x-4} times \frac{x-3}{x-3}. Multiply \frac{1}{x-3} times \frac{x-4}{x-4}.

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

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

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

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

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

Combine like terms in x-3-x+4.

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

Apply the distributive property by multiplying each term of x-4 by each term of x-3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}-3x-4x+12})$$

Combine -3x and -4x to get -7x.

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

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.

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

For any term t, t\times 1=t and 1t=t.

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