Solve (2-5y^2)-19y^3-3-4y^2+(7+6y)

$$( 2 - 5 y ^ { 2 } ) - 19 y ^ { 3 } - 3 - 4 y ^ { 2 } + ( 7 + 6 y )$$

Evaluate

$$6+6y-9y^{2}-19y^{3}$$

Solution Steps

Subtract 3 from 2 to get -1.

$$-1-5y^{2}-19y^{3}-4y^{2}+7+6y$$

Combine -5y^{2} and -4y^{2} to get -9y^{2}.

$$-1-9y^{2}-19y^{3}+7+6y$$

Add -1 and 7 to get 6.

$$6-9y^{2}-19y^{3}+6y$$

Differentiate w.r.t. y

$$6-18y-57y^{2}$$

Steps Using Derivative Rule for Sum

Subtract 3 from 2 to get -1.

$$\frac{\mathrm{d}}{\mathrm{d}y}(-1-5y^{2}-19y^{3}-4y^{2}+7+6y)$$

Combine -5y^{2} and -4y^{2} to get -9y^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(-1-9y^{2}-19y^{3}+7+6y)$$

Add -1 and 7 to get 6.

$$\frac{\mathrm{d}}{\mathrm{d}y}(6-9y^{2}-19y^{3}+6y)$$

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}.

$$2\left(-9\right)y^{2-1}+3\left(-19\right)y^{3-1}+6y^{1-1}$$

Multiply 2 times -9.

$$-18y^{2-1}+3\left(-19\right)y^{3-1}+6y^{1-1}$$

Subtract 1 from 2.

$$-18y^{1}+3\left(-19\right)y^{3-1}+6y^{1-1}$$

Multiply 3 times -19.

$$-18y^{1}-57y^{3-1}+6y^{1-1}$$

Subtract 1 from 3.

$$-18y^{1}-57y^{2}+6y^{1-1}$$

Subtract 1 from 1.

$$-18y^{1}-57y^{2}+6y^{0}$$

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

$$-18y-57y^{2}+6y^{0}$$

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

$$-18y-57y^{2}+6\times 1$$

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

$$-18y-57y^{2}+6$$