Solve sqrt{18}*sqrt{2}a

$$\sqrt { 18 } \cdot \sqrt { 2 } a$$

Evaluate

$$6a$$

Solution Steps

Factor 18=2\times 9. Rewrite the square root of the product \sqrt{2\times 9} as the product of square roots \sqrt{2}\sqrt{9}.

$$\sqrt{2}\sqrt{9}\sqrt{2}a$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2\sqrt{9}a$$

Calculate the square root of 9 and get 3.

$$2\times 3a$$

Multiply 2 and 3 to get 6.

$$6a$$

Differentiate w.r.t. a

$$6$$

Steps Using Definition of a Derivative

Factor 18=2\times 9. Rewrite the square root of the product \sqrt{2\times 9} as the product of square roots \sqrt{2}\sqrt{9}.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\sqrt{2}\sqrt{9}\sqrt{2}a)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{\mathrm{d}}{\mathrm{d}a}(2\sqrt{9}a)$$

Calculate the square root of 9 and get 3.

$$\frac{\mathrm{d}}{\mathrm{d}a}(2\times 3a)$$

Multiply 2 and 3 to get 6.

$$\frac{\mathrm{d}}{\mathrm{d}a}(6a)$$

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

$$6a^{1-1}$$

Subtract 1 from 1.

$$6a^{0}$$

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

$$6\times 1$$

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

$$6$$