Which of the following functions satisfies the equation ${f\left(x\right)}^2+{(f\text{'}(x\left)\right)}^2=1$ for all real x?

• A. $f\left(x\right)=x$
• B. $f\left(x\right)=\sin \left(x\right)$
• C. $f\left(x\right)=e^x$
• D. $f\left(x\right)=\cos \left(x\right)$

Answer: (B)

Verified step by step guidance

1. Step 1: Begin by analyzing the given equation f(x)^2 + (f'(x))^2 = 1. This equation implies that the sum of the square of the function and the square of its derivative must equal 1 for all real x.
2. Step 2: Test each candidate function by substituting it into the equation. Start with f(x) = sin(x). Compute f'(x), which is the derivative of sin(x). Using the derivative rule, f'(x) = cos(x). Substitute f(x) = sin(x) and f'(x) = cos(x) into the equation: sin(x)^2 + cos(x)^2 = 1.
3. Step 3: Recall the Pythagorean identity from trigonometry: sin(x)^2 + cos(x)^2 = 1. This confirms that f(x) = sin(x) satisfies the equation for all real x.
4. Step 4: Test the next candidate function, f(x) = cos(x). Compute f'(x), which is the derivative of cos(x). Using the derivative rule, f'(x) = -sin(x). Substitute f(x) = cos(x) and f'(x) = -sin(x) into the equation: cos(x)^2 + (-sin(x))^2 = 1.
5. Step 5: Again, use the Pythagorean identity: cos(x)^2 + sin(x)^2 = 1. This confirms that f(x) = cos(x) also satisfies the equation for all real x. The other functions, f(x) = x and f(x) = e^x, do not satisfy the equation because their derivatives and squares do not sum to 1 for all x.