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. $y=\tan{x}-\cot{x}$ $\frac{dy}{dx}=\sec^2{x}-(-\csc^2{x})$ $=\boxed{\sec^2{x}+\csc^2{x}}$