Solve (3+5c-8-5)-4*1-2+7*-3

$$( 3 + 5 c - 8 - 5 ) - 4 \cdot 1 - 2 + 7 \cdot - 3$$

Evaluate

$$5c-37$$

Solution Steps

Subtract 8 from 3 to get -5.

$$-5+5c-5-4\times 1-2+7\left(-3\right)$$

Subtract 5 from -5 to get -10.

$$-10+5c-4\times 1-2+7\left(-3\right)$$

Multiply 4 and 1 to get 4.

$$-10+5c-4-2+7\left(-3\right)$$

Subtract 4 from -10 to get -14.

$$-14+5c-2+7\left(-3\right)$$

Subtract 2 from -14 to get -16.

$$-16+5c+7\left(-3\right)$$

Multiply 7 and -3 to get -21.

$$-16+5c-21$$

Subtract 21 from -16 to get -37.

$$-37+5c$$

Differentiate w.r.t. c

$$5$$

Steps Using Derivative Rule for Sum

Subtract 8 from 3 to get -5.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-5+5c-5-4\times 1-2+7\left(-3\right))$$

Subtract 5 from -5 to get -10.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-10+5c-4\times 1-2+7\left(-3\right))$$

Multiply 4 and 1 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-10+5c-4-2+7\left(-3\right))$$

Subtract 4 from -10 to get -14.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-14+5c-2+7\left(-3\right))$$

Subtract 2 from -14 to get -16.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-16+5c+7\left(-3\right))$$

Multiply 7 and -3 to get -21.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-16+5c-21)$$

Subtract 21 from -16 to get -37.

$$\frac{\mathrm{d}}{\mathrm{d}c}(-37+5c)$$

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}.

$$5c^{1-1}$$

Subtract 1 from 1.

$$5c^{0}$$

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

$$5\times 1$$

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

$$5$$