Solve (y^{4}-6y^3+y^2+7y^3-8y-y^2-6)

$$( y ^ { 4 } - 6 y ^ { 3 } + y ^ { 2 } + 7 y ^ { 3 } - 8 y - y ^ { 2 } - 6 )$$

Evaluate

$$y^{4}+y^{3}-8y-6$$

Solution Steps

Combine -6y^{3} and 7y^{3} to get y^{3}.

$$y^{4}+y^{3}+y^{2}-8y-y^{2}-6$$

Combine y^{2} and -y^{2} to get 0.

$$y^{4}+y^{3}-8y-6$$

Differentiate w.r.t. y

$$4y^{3}+3y^{2}-8$$

Steps Using Derivative Rule for Sum

Combine -6y^{3} and 7y^{3} to get y^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y^{4}+y^{3}+y^{2}-8y-y^{2}-6)$$

Combine y^{2} and -y^{2} to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y^{4}+y^{3}-8y-6)$$

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}.

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

Subtract 1 from 4.

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

Subtract 1 from 3.

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

Subtract 1 from 1.

$$4y^{3}+3y^{2}-8y^{0}$$

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

$$4y^{3}+3y^{2}-8$$