Solve (2x+1)+(16x)+x+36x+39x

$$(2x+1)+(16x)+x+36x+39x$$

Evaluate

$$94x+1$$

Solution Steps

Combine 2x and 16x to get 18x.

$$18x+1+x+36x+39x$$

Combine 18x and x to get 19x.

$$19x+1+36x+39x$$

Combine 19x and 36x to get 55x.

$$55x+1+39x$$

Combine 55x and 39x to get 94x.

$$94x+1$$

Differentiate w.r.t. x

$$94$$

Steps Using Derivative Rule for Sum

Combine 2x and 16x to get 18x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(18x+1+x+36x+39x)$$

Combine 18x and x to get 19x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(19x+1+36x+39x)$$

Combine 19x and 36x to get 55x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(55x+1+39x)$$

Combine 55x and 39x to get 94x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(94x+1)$$

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}.

$$94x^{1-1}$$

Subtract 1 from 1.

$$94x^{0}$$

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

$$94\times 1$$

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

$$94$$