Solve 801+5+8+t+9+5+4+8+2

$$801 + 5 + 8 + t + 9 + 5 + 4 + 8 + 2$$

Evaluate

$$t+842$$

Solution Steps

Add 801 and 5 to get 806.

$$806+8+t+9+5+4+8+2$$

Add 806 and 8 to get 814.

$$814+t+9+5+4+8+2$$

Add 814 and 9 to get 823.

$$823+t+5+4+8+2$$

Add 823 and 5 to get 828.

$$828+t+4+8+2$$

Add 828 and 4 to get 832.

$$832+t+8+2$$

Add 832 and 8 to get 840.

$$840+t+2$$

Add 840 and 2 to get 842.

$$842+t$$

Differentiate w.r.t. t

$$1$$

Steps Using Derivative Rule for Sum

Add 801 and 5 to get 806.

$$\frac{\mathrm{d}}{\mathrm{d}t}(806+8+t+9+5+4+8+2)$$

Add 806 and 8 to get 814.

$$\frac{\mathrm{d}}{\mathrm{d}t}(814+t+9+5+4+8+2)$$

Add 814 and 9 to get 823.

$$\frac{\mathrm{d}}{\mathrm{d}t}(823+t+5+4+8+2)$$

Add 823 and 5 to get 828.

$$\frac{\mathrm{d}}{\mathrm{d}t}(828+t+4+8+2)$$

Add 828 and 4 to get 832.

$$\frac{\mathrm{d}}{\mathrm{d}t}(832+t+8+2)$$

Add 832 and 8 to get 840.

$$\frac{\mathrm{d}}{\mathrm{d}t}(840+t+2)$$

Add 840 and 2 to get 842.

$$\frac{\mathrm{d}}{\mathrm{d}t}(842+t)$$

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}.

$$t^{1-1}$$

Subtract 1 from 1.

$$t^{0}$$

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

$$1$$