How do you differentiate #f(x)=(t^3)(tan6t)#? Calculus Basic Differentiation Rules Product Rule 1 Answer Joel Kindiak Aug 29, 2015 Answer: If it's the correct question, #0# since differentiating constants give you #0#. If the question is #f(t)=(t^3)(tan 6t)#, then by using the chain rule, you should get #f'(t)=(3t^2)(tan 6t)[1 +(2t sec^2 6t)]#. Explanation: #f(t)=(t^3)(tan 6t)# #f'(t)=(3t^2)(tan 6t) +(t^3)(6 tan 6t sec^2 6t) # #f'(t)=(3t^2)(tan 6t) +(3t^2 tan 6t)(2t sec^2 6t) # #f'(t)=(3t^2)(tan 6t)[1 +(2t sec^2 6t)] # 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 - ... 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 140 views around the world You can reuse this answer Creative Commons License