Solve 3a4+5*4+6*9+7*8+8*6+9*3=

$$3 a 4 + 5 \cdot 4 + 6 \cdot 9 + 7 \cdot 8 + 8 \cdot 6 + 9 \cdot 3 =$$

Evaluate

$$3a_{4}+205$$

Solution Steps

Multiply 5 and 4 to get 20. Multiply 6 and 9 to get 54. Multiply 7 and 8 to get 56. Multiply 8 and 6 to get 48. Multiply 9 and 3 to get 27.

$$3a_{4}+20+54+56+48+27$$

Add 20 and 54 to get 74.

$$3a_{4}+74+56+48+27$$

Add 74 and 56 to get 130.

$$3a_{4}+130+48+27$$

Add 130 and 48 to get 178.

$$3a_{4}+178+27$$

Add 178 and 27 to get 205.

$$3a_{4}+205$$

Differentiate w.r.t. a_4

$$3$$

Steps Using Derivative Rule for Sum

Multiply 5 and 4 to get 20. Multiply 6 and 9 to get 54. Multiply 7 and 8 to get 56. Multiply 8 and 6 to get 48. Multiply 9 and 3 to get 27.

$$\frac{\mathrm{d}}{\mathrm{d}a_{4}}(3a_{4}+20+54+56+48+27)$$

Add 20 and 54 to get 74.

$$\frac{\mathrm{d}}{\mathrm{d}a_{4}}(3a_{4}+74+56+48+27)$$

Add 74 and 56 to get 130.

$$\frac{\mathrm{d}}{\mathrm{d}a_{4}}(3a_{4}+130+48+27)$$

Add 130 and 48 to get 178.

$$\frac{\mathrm{d}}{\mathrm{d}a_{4}}(3a_{4}+178+27)$$

Add 178 and 27 to get 205.

$$\frac{\mathrm{d}}{\mathrm{d}a_{4}}(3a_{4}+205)$$

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}.

$$3a_{4}^{1-1}$$

Subtract 1 from 1.

$$3a_{4}^{0}$$

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

$$3\times 1$$

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

$$3$$