Solve {l}{N^095}{-d^2-(2d-1)+2d^2-2c+2}

$$\left. \begin{array} { l } { N ^ { 0 } 95 } \\ { - d ^ { 2 } - ( 2 d - 1 ) + 2 d ^ { 2 } - 2 c + 2 } \end{array} \right.$$

Evaluate

$$98-2d+d^{2}-2c$$

Solution Steps

Calculate N to the power of 0 and get 1.

$$1\times 95-d^{2}-\left(2d-1\right)+2d^{2}-2c+2$$

Multiply 1 and 95 to get 95.

$$95-d^{2}-\left(2d-1\right)+2d^{2}-2c+2$$

To find the opposite of 2d-1, find the opposite of each term.

$$95-d^{2}-2d+1+2d^{2}-2c+2$$

Add 95 and 1 to get 96.

$$96-d^{2}-2d+2d^{2}-2c+2$$

Combine -d^{2} and 2d^{2} to get d^{2}.

$$96+d^{2}-2d-2c+2$$

Add 96 and 2 to get 98.

$$98+d^{2}-2d-2c$$

Differentiate w.r.t. d

$$2\left(d-1\right)$$