Solve (sqrt{2}+1)text{ant}(sqrt{2}-1)text{an}441

$$( \sqrt { 2 } + 1 ) \text { ant } ( \sqrt { 2 } - 1 ) \text { an } 441$$

Evaluate

$$441t\left(an\right)^{2}$$

Solution Steps

Multiply a and a to get a^{2}.

$$441\left(\sqrt{2}+1\right)a^{2}nt\left(\sqrt{2}-1\right)n$$

Multiply n and n to get n^{2}.

$$441\left(\sqrt{2}+1\right)a^{2}n^{2}t\left(\sqrt{2}-1\right)$$

Use the distributive property to multiply 441 by \sqrt{2}+1.

$$\left(441\sqrt{2}+441\right)a^{2}n^{2}t\left(\sqrt{2}-1\right)$$

Use the distributive property to multiply 441\sqrt{2}+441 by a^{2}.

$$\left(441\sqrt{2}a^{2}+441a^{2}\right)n^{2}t\left(\sqrt{2}-1\right)$$

Use the distributive property to multiply 441\sqrt{2}a^{2}+441a^{2} by n^{2}.

$$\left(441\sqrt{2}a^{2}n^{2}+441a^{2}n^{2}\right)t\left(\sqrt{2}-1\right)$$

Use the distributive property to multiply 441\sqrt{2}a^{2}n^{2}+441a^{2}n^{2} by t.

$$\left(441\sqrt{2}a^{2}n^{2}t+441a^{2}n^{2}t\right)\left(\sqrt{2}-1\right)$$

Apply the distributive property by multiplying each term of 441\sqrt{2}a^{2}n^{2}t+441a^{2}n^{2}t by each term of \sqrt{2}-1.

$$441a^{2}n^{2}t\left(\sqrt{2}\right)^{2}-441a^{2}\sqrt{2}n^{2}t+441a^{2}n^{2}t\sqrt{2}-441a^{2}n^{2}t$$

The square of \sqrt{2} is 2.

$$441a^{2}n^{2}t\times 2-441a^{2}\sqrt{2}n^{2}t+441a^{2}n^{2}t\sqrt{2}-441a^{2}n^{2}t$$

Multiply 441 and 2 to get 882.

$$882a^{2}n^{2}t-441a^{2}\sqrt{2}n^{2}t+441a^{2}n^{2}t\sqrt{2}-441a^{2}n^{2}t$$

Combine -441a^{2}\sqrt{2}n^{2}t and 441a^{2}n^{2}t\sqrt{2} to get 0.

$$882a^{2}n^{2}t-441a^{2}n^{2}t$$

Combine 882a^{2}n^{2}t and -441a^{2}n^{2}t to get 441a^{2}n^{2}t.

$$441a^{2}n^{2}t$$

Differentiate w.r.t. a

$$882atn^{2}$$