Solve d/8d^2+9d

$$\frac { d } { 8 d ^ { 2 } + 9 d }$$

Evaluate

$$\frac{1}{8d+9}$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{d}{d\left(8d+9\right)}$$

Cancel out d in both numerator and denominator.

$$\frac{1}{8d+9}$$

Differentiate w.r.t. d

$$-\frac{8}{\left(8d+9\right)^{2}}$$

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.

$$\frac{\left(8d^{2}+9d^{1}\right)\frac{\mathrm{d}}{\mathrm{d}d}(d^{1})-d^{1}\frac{\mathrm{d}}{\mathrm{d}d}(8d^{2}+9d^{1})}{\left(8d^{2}+9d^{1}\right)^{2}}$$

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}.

$$\frac{\left(8d^{2}+9d^{1}\right)d^{1-1}-d^{1}\left(2\times 8d^{2-1}+9d^{1-1}\right)}{\left(8d^{2}+9d^{1}\right)^{2}}$$

Simplify.

$$\frac{\left(8d^{2}+9d^{1}\right)d^{0}-d^{1}\left(16d^{1}+9d^{0}\right)}{\left(8d^{2}+9d^{1}\right)^{2}}$$

Multiply 8d^{2}+9d^{1} times d^{0}.

$$\frac{8d^{2}d^{0}+9d^{1}d^{0}-d^{1}\left(16d^{1}+9d^{0}\right)}{\left(8d^{2}+9d^{1}\right)^{2}}$$

Multiply d^{1} times 16d^{1}+9d^{0}.

$$\frac{8d^{2}d^{0}+9d^{1}d^{0}-\left(d^{1}\times 16d^{1}+d^{1}\times 9d^{0}\right)}{\left(8d^{2}+9d^{1}\right)^{2}}$$

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

$$\frac{8d^{2}+9d^{1}-\left(16d^{1+1}+9d^{1}\right)}{\left(8d^{2}+9d^{1}\right)^{2}}$$

Simplify.

$$\frac{8d^{2}+9d^{1}-\left(16d^{2}+9d^{1}\right)}{\left(8d^{2}+9d^{1}\right)^{2}}$$

Combine like terms.

$$\frac{-8d^{2}}{\left(8d^{2}+9d^{1}\right)^{2}}$$

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

$$\frac{-8d^{2}}{\left(8d^{2}+9d\right)^{2}}$$