Solve x+sqrt{64}+120x-74x+1200-350+72x

$$x+ \sqrt{ 64 } +120x-74x+1200-350+72x$$

Evaluate

$$119x+858$$

Solution Steps

Calculate the square root of 64 and get 8.

$$x+8+120x-74x+1200-350+72x$$

Combine x and 120x to get 121x.

$$121x+8-74x+1200-350+72x$$

Combine 121x and -74x to get 47x.

$$47x+8+1200-350+72x$$

Add 8 and 1200 to get 1208.

$$47x+1208-350+72x$$

Subtract 350 from 1208 to get 858.

$$47x+858+72x$$

Combine 47x and 72x to get 119x.

$$119x+858$$

Differentiate w.r.t. x

$$119$$

Steps Using Derivative Rule for Sum

Calculate the square root of 64 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x+8+120x-74x+1200-350+72x)$$

Combine x and 120x to get 121x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(121x+8-74x+1200-350+72x)$$

Combine 121x and -74x to get 47x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(47x+8+1200-350+72x)$$

Add 8 and 1200 to get 1208.

$$\frac{\mathrm{d}}{\mathrm{d}x}(47x+1208-350+72x)$$

Subtract 350 from 1208 to get 858.

$$\frac{\mathrm{d}}{\mathrm{d}x}(47x+858+72x)$$

Combine 47x and 72x to get 119x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(119x+858)$$

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}.

$$119x^{1-1}$$

Subtract 1 from 1.

$$119x^{0}$$

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

$$119\times 1$$

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

$$119$$