Solve 70+0.9*0.9y+320+120+40-0.31y

$$70 + 0.9 \times 0.9 y + 320 + 120 + 40 - 0.31 y$$

Evaluate

$$\frac{y}{2}+550$$

Solution Steps

Multiply 0.9 and 0.9 to get 0.81.

$$70+0.81y+320+120+40-0.31y$$

Add 70 and 320 to get 390.

$$390+0.81y+120+40-0.31y$$

Add 390 and 120 to get 510.

$$510+0.81y+40-0.31y$$

Add 510 and 40 to get 550.

$$550+0.81y-0.31y$$

Combine 0.81y and -0.31y to get 0.5y.

$$550+0.5y$$

Differentiate w.r.t. y

$$0.5$$

Steps Using Derivative Rule for Sum

Multiply 0.9 and 0.9 to get 0.81.

$$\frac{\mathrm{d}}{\mathrm{d}y}(70+0.81y+320+120+40-0.31y)$$

Add 70 and 320 to get 390.

$$\frac{\mathrm{d}}{\mathrm{d}y}(390+0.81y+120+40-0.31y)$$

Add 390 and 120 to get 510.

$$\frac{\mathrm{d}}{\mathrm{d}y}(510+0.81y+40-0.31y)$$

Add 510 and 40 to get 550.

$$\frac{\mathrm{d}}{\mathrm{d}y}(550+0.81y-0.31y)$$

Combine 0.81y and -0.31y to get 0.5y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(550+0.5y)$$

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}.

$$0.5y^{1-1}$$

Subtract 1 from 1.

$$0.5y^{0}$$

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

$$0.5\times 1$$

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

$$0.5$$