Solve (2n^4-3+6n^2)+(n-8n^2-5)

$$( 2 n ^ { 4 } - 3 + 6 n ^ { 2 } ) + ( n - 8 n ^ { 2 } - 5 )$$

Evaluate

$$2n^{4}-2n^{2}+n-8$$

Solution Steps

Combine 6n^{2} and -8n^{2} to get -2n^{2}.

$$2n^{4}-3-2n^{2}+n-5$$

Subtract 5 from -3 to get -8.

$$2n^{4}-8-2n^{2}+n$$

Differentiate w.r.t. n

$$8n^{3}-4n+1$$

Steps Using Derivative Rule for Sum

Combine 6n^{2} and -8n^{2} to get -2n^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}n}(2n^{4}-3-2n^{2}+n-5)$$

Subtract 5 from -3 to get -8.

$$\frac{\mathrm{d}}{\mathrm{d}n}(2n^{4}-8-2n^{2}+n)$$

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}.

$$4\times 2n^{4-1}+2\left(-2\right)n^{2-1}+n^{1-1}$$

Multiply 4 times 2.

$$8n^{4-1}+2\left(-2\right)n^{2-1}+n^{1-1}$$

Subtract 1 from 4.

$$8n^{3}+2\left(-2\right)n^{2-1}+n^{1-1}$$

Multiply 2 times -2.

$$8n^{3}-4n^{2-1}+n^{1-1}$$

Subtract 1 from 2.

$$8n^{3}-4n^{1}+n^{1-1}$$

Subtract 1 from 1.

$$8n^{3}-4n^{1}+n^{0}$$

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

$$8n^{3}-4n+n^{0}$$

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

$$8n^{3}-4n+1$$