ddxtan u=sec2u dudx, u 0 sec2u= 1cos2u makes a tangent derivative positive when the inner gradient is positive; asymptotes are outside the domain. Method Differentiate the angle, multiply by secant squared, and use exact cosine values for exact evaluations. Example y=tan(2x), y =2sec2(2x), y (0)=2 Exam check The derivative is secant squared, not reciprocal tangent squared. A vertical asymptote is not a point of the curve.

\(\dfrac{\mathrm{d}}{\mathrm{d}x}\tan u=\sec^2u\dfrac{\mathrm{d}u}{\mathrm{d}x},\quad\cos u\ne0\)

\(\sec^2u=\dfrac{1}{\cos^2u}\) makes a tangent derivative positive when the inner gradient is positive; asymptotes are outside the domain.

Differentiate the angle, multiply by secant squared, and use exact cosine values for exact evaluations.

\(y=\tan(2x),\quad y^{\prime}=2\sec^2(2x),\quad y^{\prime}(0)=2\)

The derivative is secant squared, not reciprocal tangent squared. A vertical asymptote is not a point of the curve.

Tangent domain and gradient

Tangent domain and gradient