Solve h4*|-15+13|

$$h 4 \cdot | - 15 + 13 |$$

Evaluate

$$2h_{4}$$

Solution Steps

Add -15 and 13 to get -2.

$$h_{4}|-2|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$h_{4}\times 2$$

Differentiate w.r.t. h_4

$$2$$

Steps Using Definition of a Derivative

Add -15 and 13 to get -2.

$$\frac{\mathrm{d}}{\mathrm{d}h_{4}}(h_{4}|-2|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$\frac{\mathrm{d}}{\mathrm{d}h_{4}}(h_{4}\times 2)$$

The derivative of ax^{n} is nax^{n-1}.

$$2h_{4}^{1-1}$$

Subtract 1 from 1.

$$2h_{4}^{0}$$

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

$$2\times 1$$

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

$$2$$