How do you differentiate ln(sec^2 * x)ln(sec2⋅x)? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer maganbhai P. Jul 5, 2018 (dy)/(dx)=2tanxdydx=2tanx Explanation: Here , y=ln(sec^2x)y=ln(sec2x) Let , y=lnu , where , u=sec^2xy=lnu,where,u=sec2x (dy)/(du)=1/u and (du)/(dx)=2secx*secxtanx=2sec^2xtanxdydu=1uanddudx=2secx⋅secxtanx=2sec2xtanx Diff.w.r.t. xx using Chain Rule: color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)dydx=dydu⋅dudx (dy)/(dx)=1/u xx 2sec^2xtanxdydx=1u×2sec2xtanx Subst, back , u=sec^2xu=sec2x :.(dy)/(dx)=1/sec^2x xx 2sec^2xtanx (dy)/(dx)=2tanx Answer link Related questions What is Derivatives of y=sec(x) ? What is the Derivative of y=sec(x^2)? What is the Derivative of y=x sec(kx)? What is the Derivative of y=sec ^ 2(x)? What is the derivative of y=4 sec ^2(x)? What is the derivative of y=ln(sec(x)+tan(x))? What is the derivative of y=sec^2(x)? What is the derivative of y=sec^2(x) + tan^2(x)? What is the derivative of y=sec^3(x)? What is the derivative of y=sec(x) tan(x)? See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x) Impact of this question 5431 views around the world You can reuse this answer Creative Commons License