Solve 1/2x^2+3x+1+1/2x^2+5x+3-3/4x^2+8x+3

$$\frac { 1 } { 2 x ^ { 2 } + 3 x + 1 } + \frac { 1 } { 2 x ^ { 2 } + 5 x + 3 } - \frac { 3 } { 4 x ^ { 2 } + 8 x + 3 }$$

Evaluate

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

Short Solution Steps

Factor 2x^{2}+3x+1. Factor 2x^{2}+5x+3.

$$\frac{1}{\left(x+1\right)\left(2x+1\right)}+\frac{1}{\left(x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3}$$

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

$$\frac{2x+3}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}+\frac{2x+1}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3}$$

Since \frac{2x+3}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)} and \frac{2x+1}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x+3+2x+1}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3}$$

Combine like terms in 2x+3+2x+1.

$$\frac{4x+4}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3}$$

Factor the expressions that are not already factored in \frac{4x+4}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}.

$$\frac{4\left(x+1\right)}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3}$$

Cancel out x+1 in both numerator and denominator.

$$\frac{4}{\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3}$$

Factor 4x^{2}+8x+3.

$$\frac{4}{\left(2x+1\right)\left(2x+3\right)}-\frac{3}{\left(2x+1\right)\left(2x+3\right)}$$

Since \frac{4}{\left(2x+1\right)\left(2x+3\right)} and \frac{3}{\left(2x+1\right)\left(2x+3\right)} have the same denominator, subtract them by subtracting their numerators. Subtract 3 from 4 to get 1.

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

Expand \left(2x+1\right)\left(2x+3\right).

$$\frac{1}{4x^{2}+8x+3}$$

Differentiate w.r.t. x

$$-\frac{8\left(x+1\right)}{\left(\left(2x+1\right)\left(2x+3\right)\right)^{2}}$$

Steps Using Chain Rule

Factor 2x^{2}+3x+1. Factor 2x^{2}+5x+3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x+1\right)\left(2x+1\right)}+\frac{1}{\left(x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x+3}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}+\frac{2x+1}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3})$$

Since \frac{2x+3}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)} and \frac{2x+1}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x+3+2x+1}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3})$$

Combine like terms in 2x+3+2x+1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4x+4}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3})$$

Factor the expressions that are not already factored in \frac{4x+4}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4\left(x+1\right)}{\left(x+1\right)\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3})$$

Cancel out x+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4}{\left(2x+1\right)\left(2x+3\right)}-\frac{3}{4x^{2}+8x+3})$$

Factor 4x^{2}+8x+3.

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

Since \frac{4}{\left(2x+1\right)\left(2x+3\right)} and \frac{3}{\left(2x+1\right)\left(2x+3\right)} have the same denominator, subtract them by subtracting their numerators. Subtract 3 from 4 to get 1.

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

Use the distributive property to multiply 2x+1 by 2x+3 and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{4x^{2}+8x+3})$$

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).

$$-\left(4x^{2}+8x^{1}+3\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(4x^{2}+8x^{1}+3)$$

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}.

$$-\left(4x^{2}+8x^{1}+3\right)^{-2}\left(2\times 4x^{2-1}+8x^{1-1}\right)$$

Simplify.

$$\left(4x^{2}+8x^{1}+3\right)^{-2}\left(-8x^{1}-8x^{0}\right)$$

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

$$\left(4x^{2}+8x+3\right)^{-2}\left(-8x-8x^{0}\right)$$

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

$$\left(4x^{2}+8x+3\right)^{-2}\left(-8x-8\right)$$