Solve x2^3-3^2*1^3+7*0=

$$x 2 ^ { 3 } - 3 ^ { 2 } \times 1 ^ { 3 } + 7 \times 0 =$$

Evaluate

$$8x-9$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$x\times 8-3^{2}\times 1^{3}+7\times 0$$

Calculate 3 to the power of 2 and get 9.

$$x\times 8-9\times 1^{3}+7\times 0$$

Calculate 1 to the power of 3 and get 1.

$$x\times 8-9\times 1+7\times 0$$

Multiply 9 and 1 to get 9.

$$x\times 8-9+7\times 0$$

Multiply 7 and 0 to get 0.

$$x\times 8-9+0$$

Add -9 and 0 to get -9.

$$x\times 8-9$$

Differentiate w.r.t. x

$$8$$

Steps Using Derivative Rule for Sum

Calculate 2 to the power of 3 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x\times 8-3^{2}\times 1^{3}+7\times 0)$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x\times 8-9\times 1^{3}+7\times 0)$$

Calculate 1 to the power of 3 and get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x\times 8-9\times 1+7\times 0)$$

Multiply 9 and 1 to get 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x\times 8-9+7\times 0)$$

Multiply 7 and 0 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x\times 8-9+0)$$

Add -9 and 0 to get -9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x\times 8-9)$$

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}.

$$8x^{1-1}$$

Subtract 1 from 1.

$$8x^{0}$$

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

$$8\times 1$$

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

$$8$$