Solve 10b/2b+8

$$\frac { 10 b } { 2 b + 8 }$$

Evaluate

$$\frac{5b}{b+4}$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{10b}{2\left(b+4\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{5b}{b+4}$$

Differentiate w.r.t. b

$$\frac{20}{\left(b+4\right)^{2}}$$

Steps Using Derivative Rule for Quotient

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.

$$\frac{\left(2b^{1}+8\right)\frac{\mathrm{d}}{\mathrm{d}b}(10b^{1})-10b^{1}\frac{\mathrm{d}}{\mathrm{d}b}(2b^{1}+8)}{\left(2b^{1}+8\right)^{2}}$$

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}.

$$\frac{\left(2b^{1}+8\right)\times 10b^{1-1}-10b^{1}\times 2b^{1-1}}{\left(2b^{1}+8\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(2b^{1}+8\right)\times 10b^{0}-10b^{1}\times 2b^{0}}{\left(2b^{1}+8\right)^{2}}$$

Expand using distributive property.

$$\frac{2b^{1}\times 10b^{0}+8\times 10b^{0}-10b^{1}\times 2b^{0}}{\left(2b^{1}+8\right)^{2}}$$

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

$$\frac{2\times 10b^{1}+8\times 10b^{0}-10\times 2b^{1}}{\left(2b^{1}+8\right)^{2}}$$

Do the arithmetic.

$$\frac{20b^{1}+80b^{0}-20b^{1}}{\left(2b^{1}+8\right)^{2}}$$

Combine like terms.

$$\frac{\left(20-20\right)b^{1}+80b^{0}}{\left(2b^{1}+8\right)^{2}}$$

Subtract 20 from 20.

$$\frac{80b^{0}}{\left(2b^{1}+8\right)^{2}}$$

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

$$\frac{80b^{0}}{\left(2b+8\right)^{2}}$$

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

$$\frac{80\times 1}{\left(2b+8\right)^{2}}$$

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

$$\frac{80}{\left(2b+8\right)^{2}}$$