Find f'(x) for f(x) = cos(2x+1): (a) with x in radians

Find $f'(x)$ for $f(x) = \cos(2x+1)$: (a) with $x$ in radians, (b) with $x$ in degrees.

Step 1: Part (a): $x$ in radians

Apply the chain rule with the outer function $\cos u$ and the inner function $u = 2x+1$: $\frac{d}{du}\cos u = -\sin u, \qquad \frac{du}{dx} = 2$.

Hence \[f'(x) = \frac{d}{du}\cos u \cdot \frac{du}{dx} = -\sin(2x+1)\cdot 2 = -2\sin(2x+1).\]

Step 2: Part (b): $x$ in degrees

The differentiation rules $(\sin u)' = \cos u$ and $(\cos u)' = -\sin u$ are valid only when the angle is measured in radians. If the argument $2x+1$ is given in degrees, it must first be converted, using $1^{\circ} = \dfrac{\pi}{180}\ \text{rad}$:

\[ f(x) = \cos\!\left(\frac{\pi}{180}\,(2x+1)\right). \]

Now apply the chain rule with $u = \dfrac{\pi}{180}(2x+1)$, for which

\[ \frac{du}{dx} = \frac{\pi}{180}\cdot 2 = \frac{\pi}{90}. \]

Therefore \[f'(x) = -\sin\!\left(\frac{\pi(2x+1)}{180}\right)\cdot\frac{\pi}{90} = -\frac{\pi}{90}\,\sin\bigl((2x+1)^{\circ}\bigr).\]

Step 3: Verification

Part (a). Differentiating once more must reproduce the known relation $y'' = -4y$ for $y = \cos(2x+1)$:

\[f''(x) = \frac{d}{dx}\bigl[-2\sin(2x+1)\bigr] = -2\cos(2x+1)\cdot 2 = -4\cos(2x+1) = -4f(x) \;\;\checkmark\]

Part (b). The two answers are genuinely different numbers, which is the whole point of the exercise. At $x = 0$:

\[\text{radians:}\quad f'(0) = -2\sin(1\ \text{rad}) \approx -1.683\]

\[\text{degrees:}\quad f'(0) = -\frac{\pi}{90}\sin(1^{\circ}) \approx -0.000609\]

The same function differentiated with respect to an argument measured in different units yields different derivatives, and the ratio of the two is exactly the conversion factor $\dfrac{\pi}{180}$ applied twice — once inside the sine and once as the extra factor. $\checkmark$