Solve (3-sqrt{x})(3+sqrt{x})

$$(3- \sqrt{ x } )(3+ \sqrt{ x } )$$

Evaluate

$$9-x$$

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}.

$$3^{2}-\left(\sqrt{x}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9-\left(\sqrt{x}\right)^{2}$$

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

$$9-x$$

Differentiate w.r.t. x

$$-1$$

Steps Using Derivative Rule for Sum

Consider \left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3^{2}-\left(\sqrt{x}\right)^{2})$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(9-\left(\sqrt{x}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(9-x)$$

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}.

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

Subtract 1 from 1.

$$-x^{0}$$

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

$$-1$$