Solve left(sqrt{1+3x}-1right)left(sqrt{1+3x}+1right)

$$\left( \sqrt{ 1+3x } -1 \right) \left( \sqrt{ 1+3x } +1 \right)$$

Evaluate

$$3x$$

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}.

$$\left(\sqrt{1+3x}\right)^{2}-1^{2}$$

Calculate \sqrt{1+3x} to the power of 2 and get 1+3x.

$$1+3x-1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$1+3x-1$$

Subtract 1 from 1 to get 0.

$$3x$$

Differentiate w.r.t. x

$$3$$

Steps Using Definition of a Derivative

Consider \left(\sqrt{1+3x}-1\right)\left(\sqrt{1+3x}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{1+3x}\right)^{2}-1^{2})$$

Calculate \sqrt{1+3x} to the power of 2 and get 1+3x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(1+3x-1^{2})$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(1+3x-1)$$

Subtract 1 from 1 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x)$$

The derivative of ax^{n} is nax^{n-1}.

$$3x^{1-1}$$

Subtract 1 from 1.

$$3x^{0}$$

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

$$3\times 1$$

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

$$3$$