Solve x/(x ^ 2 + 3x + 2) - 2/(x ^ 2 - 1) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Evaluate

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

Short Solution Steps

Factor $x^{2}+3x+2$. Factor $x^{2}-1$.

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

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

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

Since $\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}$ and $\frac{2\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}$ have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in $x\left(x-1\right)-2\left(x+2\right)$.

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

Combine like terms in $x^{2}-x-2x-4$.

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

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

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

Cancel out $x+1$ in both numerator and denominator.

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

Expand $\left(x-1\right)\left(x+2\right)$.

$$\frac{x-4}{x^{2}+x-2}$$

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Quotient

Factor $x^{2}+3x+2$. Factor $x^{2}-1$.

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

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

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

Since $\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}$ and $\frac{2\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}$ have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in $x\left(x-1\right)-2\left(x+2\right)$.

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

Combine like terms in $x^{2}-x-2x-4$.

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

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

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

Cancel out $x+1$ in both numerator and denominator.

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

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

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

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

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

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 $nax^{n-1}$.

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

Simplify.

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

Multiply $x^{2}+x^{1}-2$ times $x^{0}$.

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

Multiply $x^{1}-4$ times $2x^{1}+x^{0}$.

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

To multiply powers of the same base, add their exponents.

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

Simplify.

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

Combine like terms.

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

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

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

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

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

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

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