Solve 72:9-9x*2+72:8+4x*9

$$72 : 9 - 9 x \cdot 2 + 72 : 8 + 4 x \cdot 9$$

Evaluate

$$18x+17$$

Solution Steps

Divide 72 by 9 to get 8.

$$8-9x\times 2+\frac{72}{8}+4x\times 9$$

Multiply 9 and 2 to get 18.

$$8-18x+\frac{72}{8}+4x\times 9$$

Divide 72 by 8 to get 9.

$$8-18x+9+4x\times 9$$

Add 8 and 9 to get 17.

$$17-18x+4x\times 9$$

Multiply 4 and 9 to get 36.

$$17-18x+36x$$

Combine -18x and 36x to get 18x.

$$17+18x$$

Differentiate w.r.t. x

$$18$$

Steps Using Derivative Rule for Sum

Divide 72 by 9 to get 8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8-9x\times 2+\frac{72}{8}+4x\times 9)$$

Multiply 9 and 2 to get 18.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8-18x+\frac{72}{8}+4x\times 9)$$

Divide 72 by 8 to get 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8-18x+9+4x\times 9)$$

Add 8 and 9 to get 17.

$$\frac{\mathrm{d}}{\mathrm{d}x}(17-18x+4x\times 9)$$

Multiply 4 and 9 to get 36.

$$\frac{\mathrm{d}}{\mathrm{d}x}(17-18x+36x)$$

Combine -18x and 36x to get 18x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(17+18x)$$

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}.

$$18x^{1-1}$$

Subtract 1 from 1.

$$18x^{0}$$

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

$$18\times 1$$

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

$$18$$