Solve ( ) - 8/(d + 5) + 5/(2d - 3) | 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

$$\frac { - 8 } { d + 5 } + \frac { 5 } { 2 d - 3 }$$

Evaluate

$\frac{49-11d}{\left(2d-3\right)\left(d+5\right)}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $d+5$ and $2d-3$ is $\left(2d-3\right)\left(d+5\right)$. Multiply $\frac{-8}{d+5}$ times $\frac{2d-3}{2d-3}$. Multiply $\frac{5}{2d-3}$ times $\frac{d+5}{d+5}$.

$$\frac{-8\left(2d-3\right)}{\left(2d-3\right)\left(d+5\right)}+\frac{5\left(d+5\right)}{\left(2d-3\right)\left(d+5\right)}$$

Since $\frac{-8\left(2d-3\right)}{\left(2d-3\right)\left(d+5\right)}$ and $\frac{5\left(d+5\right)}{\left(2d-3\right)\left(d+5\right)}$ have the same denominator, add them by adding their numerators.

$$\frac{-8\left(2d-3\right)+5\left(d+5\right)}{\left(2d-3\right)\left(d+5\right)}$$

Do the multiplications in $-8\left(2d-3\right)+5\left(d+5\right)$.

$$\frac{-16d+24+5d+25}{\left(2d-3\right)\left(d+5\right)}$$

Combine like terms in $-16d+24+5d+25$.

$$\frac{-11d+49}{\left(2d-3\right)\left(d+5\right)}$$

Expand $\left(2d-3\right)\left(d+5\right)$.

$$\frac{-11d+49}{2d^{2}+7d-15}$$

Differentiate w.r.t. d

$\frac{2\left(11d^{2}-98d-89\right)}{4d^{4}+28d^{3}-11d^{2}-210d+225}$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $d+5$ and $2d-3$ is $\left(2d-3\right)\left(d+5\right)$. Multiply $\frac{-8}{d+5}$ times $\frac{2d-3}{2d-3}$. Multiply $\frac{5}{2d-3}$ times $\frac{d+5}{d+5}$.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{-8\left(2d-3\right)}{\left(2d-3\right)\left(d+5\right)}+\frac{5\left(d+5\right)}{\left(2d-3\right)\left(d+5\right)})$$

Since $\frac{-8\left(2d-3\right)}{\left(2d-3\right)\left(d+5\right)}$ and $\frac{5\left(d+5\right)}{\left(2d-3\right)\left(d+5\right)}$ have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{-8\left(2d-3\right)+5\left(d+5\right)}{\left(2d-3\right)\left(d+5\right)})$$

Do the multiplications in $-8\left(2d-3\right)+5\left(d+5\right)$.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{-16d+24+5d+25}{\left(2d-3\right)\left(d+5\right)})$$

Combine like terms in $-16d+24+5d+25$.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{-11d+49}{\left(2d-3\right)\left(d+5\right)})$$

Apply the distributive property by multiplying each term of $2d-3$ by each term of $d+5$.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{-11d+49}{2d^{2}+10d-3d-15})$$

Combine $10d$ and $-3d$ to get $7d$.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{-11d+49}{2d^{2}+7d-15})$$

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.

$$\frac{\left(2d^{2}+7d^{1}-15\right)\frac{\mathrm{d}}{\mathrm{d}d}(-11d^{1}+49)-\left(-11d^{1}+49\right)\frac{\mathrm{d}}{\mathrm{d}d}(2d^{2}+7d^{1}-15)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

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 $nax^{n-1}$.

$$\frac{\left(2d^{2}+7d^{1}-15\right)\left(-11\right)d^{1-1}-\left(-11d^{1}+49\right)\left(2\times 2d^{2-1}+7d^{1-1}\right)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

Simplify.

$$\frac{\left(2d^{2}+7d^{1}-15\right)\left(-11\right)d^{0}-\left(-11d^{1}+49\right)\left(4d^{1}+7d^{0}\right)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

Multiply $2d^{2}+7d^{1}-15$ times $-11d^{0}$.

$$\frac{2d^{2}\left(-11\right)d^{0}+7d^{1}\left(-11\right)d^{0}-15\left(-11\right)d^{0}-\left(-11d^{1}+49\right)\left(4d^{1}+7d^{0}\right)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

Multiply $-11d^{1}+49$ times $4d^{1}+7d^{0}$.

$$\frac{2d^{2}\left(-11\right)d^{0}+7d^{1}\left(-11\right)d^{0}-15\left(-11\right)d^{0}-\left(-11d^{1}\times 4d^{1}-11d^{1}\times 7d^{0}+49\times 4d^{1}+49\times 7d^{0}\right)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

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

$$\frac{2\left(-11\right)d^{2}+7\left(-11\right)d^{1}-15\left(-11\right)d^{0}-\left(-11\times 4d^{1+1}-11\times 7d^{1}+49\times 4d^{1}+49\times 7d^{0}\right)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

Simplify.

$$\frac{-22d^{2}-77d^{1}+165d^{0}-\left(-44d^{2}-77d^{1}+196d^{1}+343d^{0}\right)}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

Combine like terms.

$$\frac{22d^{2}-196d^{1}-178d^{0}}{\left(2d^{2}+7d^{1}-15\right)^{2}}$$

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

$$\frac{22d^{2}-196d-178d^{0}}{\left(2d^{2}+7d-15\right)^{2}}$$

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

$$\frac{22d^{2}-196d-178}{\left(2d^{2}+7d-15\right)^{2}}$$