Solve {l}{4(-2)-|-2-3|}{-8-5h}{-13}

$$\left. \begin{array} { l } { 4 ( - 2 ) - | - 2 - 3 | } \\ { - 8 - 5 h } \\ { - 13 } \end{array} \right.$$

Evaluate

$$-5h-34$$

Solution Steps

Multiply 4 and -2 to get -8.

$$-8-|-2-3|-8-5h-13$$

Subtract 3 from -2 to get -5.

$$-8-|-5|-8-5h-13$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -5 is 5.

$$-8-5-8-5h-13$$

Subtract 5 from -8 to get -13.

$$-13-8-5h-13$$

Subtract 8 from -13 to get -21.

$$-21-5h-13$$

Subtract 13 from -21 to get -34.

$$-34-5h$$

Differentiate w.r.t. h

$$-5$$

Steps Using Derivative Rule for Sum

Multiply 4 and -2 to get -8.

$$\frac{\mathrm{d}}{\mathrm{d}h}(-8-|-2-3|-8-5h-13)$$

Subtract 3 from -2 to get -5.

$$\frac{\mathrm{d}}{\mathrm{d}h}(-8-|-5|-8-5h-13)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -5 is 5.

$$\frac{\mathrm{d}}{\mathrm{d}h}(-8-5-8-5h-13)$$

Subtract 5 from -8 to get -13.

$$\frac{\mathrm{d}}{\mathrm{d}h}(-13-8-5h-13)$$

Subtract 8 from -13 to get -21.

$$\frac{\mathrm{d}}{\mathrm{d}h}(-21-5h-13)$$

Subtract 13 from -21 to get -34.

$$\frac{\mathrm{d}}{\mathrm{d}h}(-34-5h)$$

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}.

$$-5h^{1-1}$$

Subtract 1 from 1.

$$-5h^{0}$$

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

$$-5$$