Solve (sqrt{5}+sqrt{2x})*(sqrt{5}-sqrt{2x})

$$( \sqrt { 5 } + \sqrt { 2 x } ) \cdot ( \sqrt { 5 } - \sqrt { 2 x } )$$

Evaluate

$$5-2x$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\sqrt{5}\right)^{2}-\left(\sqrt{2x}\right)^{2}$$

The square of \sqrt{5} is 5.

$$5-\left(\sqrt{2x}\right)^{2}$$

Calculate \sqrt{2x} to the power of 2 and get 2x.

$$5-2x$$

Differentiate w.r.t. x

$$-2$$

Steps Using Derivative Rule for Sum

Consider \left(\sqrt{5}+\sqrt{2x}\right)\left(\sqrt{5}-\sqrt{2x}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{5}\right)^{2}-\left(\sqrt{2x}\right)^{2})$$

The square of \sqrt{5} is 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(5-\left(\sqrt{2x}\right)^{2})$$

Calculate \sqrt{2x} to the power of 2 and get 2x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(5-2x)$$

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}.

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

Subtract 1 from 1.

$$-2x^{0}$$

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

$$-2$$