Solve left(sqrt{9+x}-3right)left(sqrt{9+x}+3right)

$$\left( \sqrt{ 9+x } -3 \right) \left( \sqrt{ 9+x } +3 \right)$$

Evaluate

$$x$$

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}.

$$\left(\sqrt{9+x}\right)^{2}-3^{2}$$

Calculate \sqrt{9+x} to the power of 2 and get 9+x.

$$9+x-3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9+x-9$$

Subtract 9 from 9 to get 0.

$$x$$

Differentiate w.r.t. x

$$1$$

Steps Using Definition of a Derivative

Consider \left(\sqrt{9+x}-3\right)\left(\sqrt{9+x}+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{9+x}\right)^{2}-3^{2})$$

Calculate \sqrt{9+x} to the power of 2 and get 9+x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(9+x-3^{2})$$

Calculate 3 to the power of 2 and get 9.

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

Subtract 9 from 9 to get 0.

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

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.

$$1$$