Solve 4quady^4+1+2y^2=

$$4 \quad y ^ { 4 } + 1 + 2 y ^ { 2 } =$$

Differentiate w.r.t. y

$$16y^{3}+4y$$

Steps Using Derivative Rule for Sum

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}.

$$4\times 4y^{4-1}+2\times 2y^{2-1}$$

Multiply 4 times 4.

$$16y^{4-1}+2\times 2y^{2-1}$$

Subtract 1 from 4.

$$16y^{3}+2\times 2y^{2-1}$$

Multiply 2 times 2.

$$16y^{3}+4y^{2-1}$$

Subtract 1 from 2.

$$16y^{3}+4y^{1}$$

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

$$16y^{3}+4y$$

Evaluate

$$4y^{4}+2y^{2}+1$$