Solve (4)/(x-8(x)^(2))((x)^(2))/()-(x)/()-(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

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

Evaluate

$\frac{8x^{2}+11x-1}{1-8x}$

Short Solution Steps

Divide $1$ by $1$ to get $1$.

$$\frac{4}{x-8x^{2}}\times \frac{x^{2}}{1}-\frac{x}{1}-1$$

Anything divided by one gives itself.

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

Express $\frac{4}{x-8x^{2}}x^{2}$ as a single fraction.

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

Anything divided by one gives itself.

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

Factor $x-8x^{2}$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $-x-1$ times $\frac{x\left(-8x+1\right)}{x\left(-8x+1\right)}$.

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

Since $\frac{4x^{2}}{x\left(-8x+1\right)}$ and $\frac{\left(-x-1\right)x\left(-8x+1\right)}{x\left(-8x+1\right)}$ have the same denominator, add them by adding their numerators.

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

Do the multiplications in $4x^{2}+\left(-x-1\right)x\left(-8x+1\right)$.

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

Combine like terms in $4x^{2}+8x^{3}-x^{2}+8x^{2}-x$.

$$\frac{11x^{2}+8x^{3}-x}{x\left(-8x+1\right)}$$

Factor the expressions that are not already factored in $\frac{11x^{2}+8x^{3}-x}{x\left(-8x+1\right)}$.

$$\frac{8x\left(x-\left(-\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)\left(x-\left(\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)}{x\left(-8x+1\right)}$$

Cancel out $x$ in both numerator and denominator.

$$\frac{8\left(x-\left(-\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)\left(x-\left(\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)}{-8x+1}$$

To find the opposite of $-\frac{3}{16}\sqrt{17}-\frac{11}{16}$, find the opposite of each term.

$$\frac{8\left(x+\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)\left(x-\left(\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)}{-8x+1}$$

To find the opposite of $\frac{3}{16}\sqrt{17}-\frac{11}{16}$, find the opposite of each term.

$$\frac{8\left(x+\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)\left(x-\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)}{-8x+1}$$

Use the distributive property to multiply $8$ by $x+\frac{3}{16}\sqrt{17}+\frac{11}{16}$.

$$\frac{\left(8x+\frac{3}{2}\sqrt{17}+\frac{11}{2}\right)\left(x-\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)}{-8x+1}$$

Use the distributive property to multiply $8x+\frac{3}{2}\sqrt{17}+\frac{11}{2}$ by $x-\frac{3}{16}\sqrt{17}+\frac{11}{16}$ and combine like terms.

$$\frac{8x^{2}+11x-\frac{9}{32}\left(\sqrt{17}\right)^{2}+\frac{121}{32}}{-8x+1}$$

The square of $\sqrt{17}$ is $17$.

$$\frac{8x^{2}+11x-\frac{9}{32}\times 17+\frac{121}{32}}{-8x+1}$$

Multiply $-\frac{9}{32}$ and $17$ to get $-\frac{153}{32}$.

$$\frac{8x^{2}+11x-\frac{153}{32}+\frac{121}{32}}{-8x+1}$$

Add $-\frac{153}{32}$ and $\frac{121}{32}$ to get $-1$.

$$\frac{8x^{2}+11x-1}{-8x+1}$$

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Quotient

Divide $1$ by $1$ to get $1$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4}{x-8x^{2}}\times \frac{x^{2}}{1}-\frac{x}{1}-1)$$

Anything divided by one gives itself.

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

Express $\frac{4}{x-8x^{2}}x^{2}$ as a single fraction.

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

Anything divided by one gives itself.

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

Factor $x-8x^{2}$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $-x-1$ times $\frac{x\left(-8x+1\right)}{x\left(-8x+1\right)}$.

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

Since $\frac{4x^{2}}{x\left(-8x+1\right)}$ and $\frac{\left(-x-1\right)x\left(-8x+1\right)}{x\left(-8x+1\right)}$ have the same denominator, add them by adding their numerators.

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

Do the multiplications in $4x^{2}+\left(-x-1\right)x\left(-8x+1\right)$.

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

Combine like terms in $4x^{2}+8x^{3}-x^{2}+8x^{2}-x$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{11x^{2}+8x^{3}-x}{x\left(-8x+1\right)})$$

Factor the expressions that are not already factored in $\frac{11x^{2}+8x^{3}-x}{x\left(-8x+1\right)}$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x\left(x-\left(-\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)\left(x-\left(\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)}{x\left(-8x+1\right)})$$

Cancel out $x$ in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8\left(x-\left(-\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)\left(x-\left(\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)}{-8x+1})$$

To find the opposite of $-\frac{3}{16}\sqrt{17}-\frac{11}{16}$, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8\left(x+\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)\left(x-\left(\frac{3}{16}\sqrt{17}-\frac{11}{16}\right)\right)}{-8x+1})$$

To find the opposite of $\frac{3}{16}\sqrt{17}-\frac{11}{16}$, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8\left(x+\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)\left(x-\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)}{-8x+1})$$

Use the distributive property to multiply $8$ by $x+\frac{3}{16}\sqrt{17}+\frac{11}{16}$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\left(8x+\frac{3}{2}\sqrt{17}+\frac{11}{2}\right)\left(x-\frac{3}{16}\sqrt{17}+\frac{11}{16}\right)}{-8x+1})$$

Use the distributive property to multiply $8x+\frac{3}{2}\sqrt{17}+\frac{11}{2}$ by $x-\frac{3}{16}\sqrt{17}+\frac{11}{16}$ and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x^{2}+11x-\frac{9}{32}\left(\sqrt{17}\right)^{2}+\frac{121}{32}}{-8x+1})$$

The square of $\sqrt{17}$ is $17$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x^{2}+11x-\frac{9}{32}\times 17+\frac{121}{32}}{-8x+1})$$

Multiply $-\frac{9}{32}$ and $17$ to get $-\frac{153}{32}$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x^{2}+11x-\frac{153}{32}+\frac{121}{32}}{-8x+1})$$

Add $-\frac{153}{32}$ and $\frac{121}{32}$ to get $-1$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x^{2}+11x-1}{-8x+1})$$

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.

$$\frac{\left(-8x^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(8x^{2}+11x^{1}-1)-\left(8x^{2}+11x^{1}-1\right)\frac{\mathrm{d}}{\mathrm{d}x}(-8x^{1}+1)}{\left(-8x^{1}+1\right)^{2}}$$

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 $nax^{n-1}$.

$$\frac{\left(-8x^{1}+1\right)\left(2\times 8x^{2-1}+11x^{1-1}\right)-\left(8x^{2}+11x^{1}-1\right)\left(-8\right)x^{1-1}}{\left(-8x^{1}+1\right)^{2}}$$

Simplify.

$$\frac{\left(-8x^{1}+1\right)\left(16x^{1}+11x^{0}\right)-\left(8x^{2}+11x^{1}-1\right)\left(-8\right)x^{0}}{\left(-8x^{1}+1\right)^{2}}$$

Multiply $-8x^{1}+1$ times $16x^{1}+11x^{0}$.

$$\frac{-8x^{1}\times 16x^{1}-8x^{1}\times 11x^{0}+16x^{1}+11x^{0}-\left(8x^{2}+11x^{1}-1\right)\left(-8\right)x^{0}}{\left(-8x^{1}+1\right)^{2}}$$

Multiply $8x^{2}+11x^{1}-1$ times $-8x^{0}$.

$$\frac{-8x^{1}\times 16x^{1}-8x^{1}\times 11x^{0}+16x^{1}+11x^{0}-\left(8x^{2}\left(-8\right)x^{0}+11x^{1}\left(-8\right)x^{0}-\left(-8x^{0}\right)\right)}{\left(-8x^{1}+1\right)^{2}}$$

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

$$\frac{-8\times 16x^{1+1}-8\times 11x^{1}+16x^{1}+11x^{0}-\left(8\left(-8\right)x^{2}+11\left(-8\right)x^{1}-\left(-8x^{0}\right)\right)}{\left(-8x^{1}+1\right)^{2}}$$

Simplify.

$$\frac{-128x^{2}-88x^{1}+16x^{1}+11x^{0}-\left(-64x^{2}-88x^{1}+8x^{0}\right)}{\left(-8x^{1}+1\right)^{2}}$$

Combine like terms.

$$\frac{-64x^{2}+16x^{1}+3x^{0}}{\left(-8x^{1}+1\right)^{2}}$$

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

$$\frac{-64x^{2}+16x+3x^{0}}{\left(-8x+1\right)^{2}}$$

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

$$\frac{-64x^{2}+16x+3\times 1}{\left(-8x+1\right)^{2}}$$

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

$$\frac{-64x^{2}+16x+3}{\left(-8x+1\right)^{2}}$$