Solve 3x^2-1*(-5)-08*(-6)

$$3 x ^ { 2 } - 1 \cdot ( - 5 ) - 08 \cdot ( - 6 )$$

Evaluate

$$3x^{2}+5$$

Solution Steps

Multiply 1 and -5 to get -5.

$$3x^{2}-\left(-5\right)-0\times 8\left(-6\right)$$

The opposite of -5 is 5.

$$3x^{2}+5-0\times 8\left(-6\right)$$

Multiply 0 and 8 to get 0.

$$3x^{2}+5-0\left(-6\right)$$

Multiply 0 and -6 to get 0.

$$3x^{2}+5-0$$

Multiply -1 and 0 to get 0.

$$3x^{2}+5+0$$

Add 5 and 0 to get 5.

$$3x^{2}+5$$

Differentiate w.r.t. x

$$6x$$

Steps Using Derivative Rule for Sum

Multiply 1 and -5 to get -5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}-\left(-5\right)-0\times 8\left(-6\right))$$

The opposite of -5 is 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}+5-0\times 8\left(-6\right))$$

Multiply 0 and 8 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}+5-0\left(-6\right))$$

Multiply 0 and -6 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}+5-0)$$

Multiply -1 and 0 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}+5+0)$$

Add 5 and 0 to get 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}+5)$$

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}.

$$2\times 3x^{2-1}$$

Multiply 2 times 3.

$$6x^{2-1}$$

Subtract 1 from 2.

$$6x^{1}$$

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

$$6x$$