Solve 2/3x^2+16x+21+1/3x^2+13x+14

$$\frac { 2 } { 3 x ^ { 2 } + 16 x + 21 } + \frac { 1 } { 3 x ^ { 2 } + 13 x + 14 }$$

Evaluate

$$\frac{1}{\left(x+2\right)\left(x+3\right)}$$

Short Solution Steps

Factor 3x^{2}+16x+21. Factor 3x^{2}+13x+14.

$$\frac{2}{\left(x+3\right)\left(3x+7\right)}+\frac{1}{\left(x+2\right)\left(3x+7\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+3\right)\left(3x+7\right) and \left(x+2\right)\left(3x+7\right) is \left(x+2\right)\left(x+3\right)\left(3x+7\right). Multiply \frac{2}{\left(x+3\right)\left(3x+7\right)} times \frac{x+2}{x+2}. Multiply \frac{1}{\left(x+2\right)\left(3x+7\right)} times \frac{x+3}{x+3}.

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

Since \frac{2\left(x+2\right)}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)} and \frac{x+3}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(x+2\right)+x+3.

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

Combine like terms in 2x+4+x+3.

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

Cancel out 3x+7 in both numerator and denominator.

$$\frac{1}{\left(x+2\right)\left(x+3\right)}$$

Expand \left(x+2\right)\left(x+3\right).

$$\frac{1}{x^{2}+5x+6}$$

Differentiate w.r.t. x

$$\frac{-2x-5}{\left(\left(x+2\right)\left(x+3\right)\right)^{2}}$$

Steps Using Chain Rule

Factor 3x^{2}+16x+21. Factor 3x^{2}+13x+14.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2}{\left(x+3\right)\left(3x+7\right)}+\frac{1}{\left(x+2\right)\left(3x+7\right)})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+3\right)\left(3x+7\right) and \left(x+2\right)\left(3x+7\right) is \left(x+2\right)\left(x+3\right)\left(3x+7\right). Multiply \frac{2}{\left(x+3\right)\left(3x+7\right)} times \frac{x+2}{x+2}. Multiply \frac{1}{\left(x+2\right)\left(3x+7\right)} times \frac{x+3}{x+3}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2\left(x+2\right)}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)}+\frac{x+3}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)})$$

Since \frac{2\left(x+2\right)}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)} and \frac{x+3}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2\left(x+2\right)+x+3}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)})$$

Do the multiplications in 2\left(x+2\right)+x+3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x+4+x+3}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)})$$

Combine like terms in 2x+4+x+3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x+7}{\left(x+2\right)\left(x+3\right)\left(3x+7\right)})$$

Cancel out 3x+7 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x+2\right)\left(x+3\right)})$$

Use the distributive property to multiply x+2 by x+3 and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+5x+6})$$

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).

$$-\left(x^{2}+5x^{1}+6\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+5x^{1}+6)$$

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}.

$$-\left(x^{2}+5x^{1}+6\right)^{-2}\left(2x^{2-1}+5x^{1-1}\right)$$

Simplify.

$$\left(x^{2}+5x^{1}+6\right)^{-2}\left(-2x^{1}-5x^{0}\right)$$

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

$$\left(x^{2}+5x+6\right)^{-2}\left(-2x-5x^{0}\right)$$

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

$$\left(x^{2}+5x+6\right)^{-2}\left(-2x-5\right)$$