How do you find the derivative of # (pi/2)sinx-cosx#? Calculus Basic Differentiation Rules Summary of Differentiation Rules 1 Answer Alan N. Jun 24, 2018 #dy/dx = (pi/2)cosx + sinx# Explanation: #y = (pi/2)sinx - cosx# Since #pi/2# is a constant we can apply linearity for the derivative. Hence, #dy/dx = (pi/2)d/dx(sinx) - d/dx(cosx)# Applying standard derivatives #dy/dx = (pi/2)cosx - (-sinx)# #= (pi/2)cosx + sinx# Answer link Related questions What is a summary of Differentiation Rules? What are the first three derivatives of #(xcos(x)-sin(x))/(x^2)#? How do you find the derivative of #(e^(2x) - e^(-2x))/(e^(2x) + e^(-2x))#? How do I find the derivative of #y= x arctan (2x) - (ln (1+4x^2))/4#? How do you find the derivative of #y = s/3 + 5s#? What is the second derivative of #(f * g)(x)# if f and g are functions such that #f'(x)=g(x)#... How do you calculate the derivative for #g(t)= 7/sqrtt#? Can you use a calculator to differentiate #f(x) = 3x^2 + 12#? What is the derivative of #ln(x)+ 3 ln(x) + 5/7x +(2/x)#? How do you find the formula for the derivative of #1/x#? See all questions in Summary of Differentiation Rules Impact of this question 6060 views around the world You can reuse this answer Creative Commons License