Solve 1^2+2^2+3^2+1*1+n^2

$$1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + 1 \cdot 1 + n ^ { 2 }$$

Evaluate

$$n^{2}+15$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$1+2^{2}+3^{2}+1\times 1+n^{2}$$

Calculate 2 to the power of 2 and get 4.

$$1+4+3^{2}+1\times 1+n^{2}$$

Add 1 and 4 to get 5.

$$5+3^{2}+1\times 1+n^{2}$$

Calculate 3 to the power of 2 and get 9.

$$5+9+1\times 1+n^{2}$$

Add 5 and 9 to get 14.

$$14+1\times 1+n^{2}$$

Multiply 1 and 1 to get 1.

$$14+1+n^{2}$$

Add 14 and 1 to get 15.

$$15+n^{2}$$

Differentiate w.r.t. n

$$2n$$

Steps Using Derivative Rule for Sum

Calculate 1 to the power of 2 and get 1.

$$\frac{\mathrm{d}}{\mathrm{d}n}(1+2^{2}+3^{2}+1\times 1+n^{2})$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\mathrm{d}}{\mathrm{d}n}(1+4+3^{2}+1\times 1+n^{2})$$

Add 1 and 4 to get 5.

$$\frac{\mathrm{d}}{\mathrm{d}n}(5+3^{2}+1\times 1+n^{2})$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\mathrm{d}}{\mathrm{d}n}(5+9+1\times 1+n^{2})$$

Add 5 and 9 to get 14.

$$\frac{\mathrm{d}}{\mathrm{d}n}(14+1\times 1+n^{2})$$

Multiply 1 and 1 to get 1.

$$\frac{\mathrm{d}}{\mathrm{d}n}(14+1+n^{2})$$

Add 14 and 1 to get 15.

$$\frac{\mathrm{d}}{\mathrm{d}n}(15+n^{2})$$

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}.

$$2n^{2-1}$$

Subtract 1 from 2.

$$2n^{1}$$

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

$$2n$$