How do you find the derivative of #sin^2x+cos^2x#? Calculus Differentiating Trigonometric Functions Differentiating sin(x) from First Principles 1 Answer MeneerNask May 15, 2015 This must be a trick question. #sin^2x+cos^2x=1# (allways), so the derivative is #0# Answer link Related questions How do you differentiate #f(x)=sin(x)# from first principles? What is the derivative of #y=3sin(x) - sin(3x)#? How do you find dy/dx if #x + tan(xy) = 0#? How do you find the derivative of the function #y=cos((1-e^(2x))/(1+e^(2x)))#? How do you differentiate #f(x)=2secx+(2e^x)(tanx)#? How do you find the derivate for #y = pisinx - 4cosx#? How do you find the derivative of #f(t) = t^2sin t#? What is the derivative of #sin^2(lnx)#? How do you compute the 200th derivative of #f(x)=sin(2x)#? How do you find the derivative of #sin(x^2+1)#? See all questions in Differentiating sin(x) from First Principles Impact of this question 2097 views around the world You can reuse this answer Creative Commons License