Question

Give me differentiation of sin x/cos2x

Answer 1

f'(x)=\frac{\cos x\cos2x+2\sin x\sin2x}{\cos^2 2x}.

Answer 2

Solution

Let

$$f(x)=\frac{\sin x}{\cos2x}.$$

Using the quotient rule:

$$f'(x)=\frac{(\sin x)' \cdot \cos 2x - \sin x \cdot (\cos2x)'}{(\cos2x)^2}.$$

We have:

$$(\sin x)'=\cos x,$$

and

$$(\cos2x)'=-2\sin2x.$$

Thus,

$$f'(x)=\frac{\cos x\cdot\cos2x - \sin x \cdot (-2\sin2x)}{(\cos2x)^2}=\frac{\cos x\cdot\cos2x+2\sin x\sin2x}{(\cos2x)^2}.$$

Explanation (minimal): Differentiate using the quotient rule, noting that $(\sin x)'=\cos x$ and $(\cos2x)'=-2\sin2x$. Substitute these into the formula and simplify.