Solve (sqrt{2}x-1)(sqrt{2}x-6)

$$( \sqrt { 2 } x - 1 ) ( \sqrt { 2 } x - 6 )$$

Evaluate

$$\left(\sqrt{2}x-6\right)\left(\sqrt{2}x-1\right)$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{2}x-1 by each term of \sqrt{2}x-6.

$$\left(\sqrt{2}\right)^{2}x^{2}-6\sqrt{2}x-\sqrt{2}x+6$$

The square of \sqrt{2} is 2.

$$2x^{2}-6\sqrt{2}x-\sqrt{2}x+6$$

Combine -6\sqrt{2}x and -\sqrt{2}x to get -7\sqrt{2}x.

$$2x^{2}-7\sqrt{2}x+6$$

Differentiate w.r.t. x

$$4x-7\sqrt{2}$$

Steps Using Derivative Rule for Sum

Apply the distributive property by multiplying each term of \sqrt{2}x-1 by each term of \sqrt{2}x-6.

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

The square of \sqrt{2} is 2.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-6\sqrt{2}x-\sqrt{2}x+6)$$

Combine -6\sqrt{2}x and -\sqrt{2}x to get -7\sqrt{2}x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-7\sqrt{2}x+6)$$

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}.

$$2\times 2x^{2-1}+\left(-7\sqrt{2}\right)x^{1-1}$$

Multiply 2 times 2.

$$4x^{2-1}+\left(-7\sqrt{2}\right)x^{1-1}$$

Subtract 1 from 2.

$$4x^{1}+\left(-7\sqrt{2}\right)x^{1-1}$$

Subtract 1 from 1.

$$4x^{1}+\left(-7\sqrt{2}\right)x^{0}$$

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

$$4x+\left(-7\sqrt{2}\right)x^{0}$$

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

$$4x+\left(-7\sqrt{2}\right)\times 1$$

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

$$4x-7\sqrt{2}$$