How do you differentiate #y = 3x cos^2 (x)#? Calculus Basic Differentiation Rules Product Rule 1 Answer Barry H. Feb 1, 2018 #3cosx#[#cosx-2xsinx#] Explanation: #d#[#uv#]= #vdu+udv# this is the product rule, where v and u are both functions of x. Let u =#3x# and #v=cos^2x# so we have ......#cos^2x#.[#3]#+#3x#[#-2sinxcosx#] =#3cos^2x#-#6xsinxcosx#.....=#3cosx[cosx-2xsinx#]. #cos^2x#=#[cosx]^2# and so need to use the chain rule to differentiate this, =#2cosx#, times the derivative of #cosx# which is -#sinx#, so# d/dxcos^2x##=-2sinxcosx#. hope this helped. Answer link Related questions What is the Product Rule for derivatives? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x - 3)(2 - 3x)(5 - x)# ? How do you use the product rule to find the derivative of #y=x^2*sin(x)# ? How do you use the product rule to differentiate #y=cos(x)*sin(x)# ? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x^4 +x)*e^x*tan(x)# ? How do you use the product rule to find the derivative of #y=(x^3+2x)*e^x# ? How do you use the product rule to find the derivative of #y=sqrt(x)*cos(x)# ? How do you use the product rule to find the derivative of #y=(1/x^2-3/x^4)*(x+5x^3)# ? How do you use the product rule to find the derivative of #y=sqrt(x)*e^x# ? How do you use the product rule to find the derivative of #y=x*ln(x)# ? See all questions in Product Rule Impact of this question 4483 views around the world You can reuse this answer Creative Commons License