Solve 73quadxquad45

$$73 \quad x \quad 45$$

Evaluate

$$73x_{45}$$

Steps For Long Multiplication

First line up the numbers vertically and match the places from the right like this.

$$\begin{array}{c}\phantom{\times99}73\\\underline{\times\phantom{99}45}\\\end{array}$$

Now multiply the first number with the 1^{st} digit in 2^{nd} number to get intermediate results. That is Multiply 73 with 5. Write the result 365 at the end leaving 0 spaces to the right like this.

$$\begin{array}{c}\phantom{\times99}73\\\underline{\times\phantom{99}45}\\\phantom{\times9}365\\\end{array}$$

Now multiply the first number with the 2^{nd} digit in 2^{nd} number to get intermediate results. That is Multiply 73 with 4. Write the result 292 at the end leaving 1 spaces to the right like this.

$$\begin{array}{c}\phantom{\times99}73\\\underline{\times\phantom{99}45}\\\phantom{\times9}365\\\underline{\phantom{\times}292\phantom{9}}\\\end{array}$$

Now add the intermediate results to get final answer.

$$\begin{array}{c}\phantom{\times99}73\\\underline{\times\phantom{99}45}\\\phantom{\times9}365\\\underline{\phantom{\times}292\phantom{9}}\\\phantom{\times}3285\end{array}$$

Differentiate w.r.t. x_45

$$73$$

Steps Using Definition of a Derivative

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

$$73x_{45}^{1-1}$$

Subtract 1 from 1.

$$73x_{45}^{0}$$

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

$$73\times 1$$

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

$$73$$