Differentiate #sinx/(5x+sec^2x)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer VinÃcius Ferraz Jun 7, 2017 #y' = frac{cos x (5x + sec^2 x) - sin x (5 + 2 sec x frac{1}{cos^2 x} sin x)}{(5x + sec^2x)^2}# Explanation: #D_x y = frac{cos x (5x + sec^2 x) - sin x (5 + 2 sec x D_x frac{1}{cos x})}{(5x + sec^2x)^2}# 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 1714 views around the world You can reuse this answer Creative Commons License