Differentiate #sinx# #/# #5x# + #sec^2 x"# ? Calculus Derivatives Differentiable vs. Non-differentiable Functions 1 Answer Justin Jun 6, 2017 #(xcosx-sinx)/(5x^2)+2sec^2xtanx# Explanation: #d/dx(sinx/(5x)+sec^2x)# #=d/dx(sinx/(5x))+d/dx(sec^2x)# #=((d(sinx))/dx(5x)-sinx(d(5x))/dx)/(5x)^2+(d(sec^2x))/(dsecx)*(d(secx))/dx# #=(cosx*5x-5sinx)/(25x^2)+2secx*secxtanx# Answer link Related questions What are non differentiable points for a function? What does differentiable mean for a function? What are non differentiable points for a graph? How do you find the non differentiable points for a function? How do you find the non differentiable points for a graph? What are differentiable points for a function? How do you find the differentiable points for a graph? What is the derivative of a unit vector? On what interval is the function #ln((4x^2)+9)# differentiable? How do you find the partial derivative of the function #f(x,y)=intcos(-7t^2-6t-1)dt#? See all questions in Differentiable vs. Non-differentiable Functions Impact of this question 2139 views around the world You can reuse this answer Creative Commons License