Differentiation Question 3

If $y=\tan{x}-\cot{x}$, then $ \dfrac{dy}{dx}=$

$ \sec{x}\csc{x}$

$ \sec{x}+\csc{x}$

$ \sec^2{x}-\csc^2{x}$

$ \sec^2{x}+\csc^2{x}$

Approach

Derive using the rules for tangent and cotangent. $$y=\tan{x}-\cot{x}$$ $$\frac{dy}{dx}=\sec^2{x}-(-\csc^2{x})$$ $$ =\boxed{\sec^2{x}+\csc^2{x}} $$