Solve 2ef^2-2(2ef)(1)+1^2

$$2 e f ^ { 2 } - 2 ( 2 e f ) ( 1 ) + 1 ^ { 2 }$$

Evaluate

$$2ef^{2}-4ef+1$$

Solution Steps

Multiply 2 and 2 to get 4.

$$2ef^{2}-4ef\times 1+1^{2}$$

Multiply 4 and 1 to get 4.

$$2ef^{2}-4ef+1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$2ef^{2}-4ef+1$$

Differentiate w.r.t. f

$$4e\left(f-1\right)$$

Steps Using Derivative Rule for Sum

Multiply 2 and 2 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}f}(2ef^{2}-4ef\times 1+1^{2})$$

Multiply 4 and 1 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}f}(2ef^{2}-4ef+1^{2})$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\mathrm{d}}{\mathrm{d}f}(2ef^{2}-4ef+1)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2\times 2ef^{2-1}+\left(-4e\right)f^{1-1}$$

Multiply 2 times 2e.

$$4ef^{2-1}+\left(-4e\right)f^{1-1}$$

Subtract 1 from 2.

$$4ef^{1}+\left(-4e\right)f^{1-1}$$

Subtract 1 from 1.

$$4ef^{1}+\left(-4e\right)f^{0}$$

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

$$4ef+\left(-4e\right)f^{0}$$

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

$$4ef+\left(-4e\right)\times 1$$

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

$$4ef-4e$$