Solve (sqrt{(2-x)^2+25})^2

$$( \sqrt { ( 2 - x ) ^ { 2 } + 25 } ) ^ { 2 }$$

Evaluate

$$\left(2-x\right)^{2}+25$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.

$$\left(\sqrt{4-4x+x^{2}+25}\right)^{2}$$

Add 4 and 25 to get 29.

$$\left(\sqrt{29-4x+x^{2}}\right)^{2}$$

Calculate \sqrt{29-4x+x^{2}} to the power of 2 and get 29-4x+x^{2}.

$$29-4x+x^{2}$$

Differentiate w.r.t. x

$$2\left(x-2\right)$$

Steps Using Derivative Rule for Sum

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{4-4x+x^{2}+25}\right)^{2})$$

Add 4 and 25 to get 29.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{29-4x+x^{2}}\right)^{2})$$

Calculate \sqrt{29-4x+x^{2}} to the power of 2 and get 29-4x+x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(29-4x+x^{2})$$

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}.

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

Subtract 1 from 1.

$$-4x^{0}+2x^{2-1}$$

Subtract 1 from 2.

$$-4x^{0}+2x^{1}$$

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

$$-4x^{0}+2x$$

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

$$-4+2x$$