How do you differentiate #f(t)=sin^2(e^(sin^2t))# using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Barry H. Feb 2, 2018 #2sine^[sin^2t]##cose^[sin^2t]# Explanation: #sin^2[e^[sin^2t]]= [sine^[Sin^2t]]^2# so using the chain rule , we have .... #2[sine^[sin^2t]]# times the differential within the the brackets ie...... #cos[e^[sin^2t]]# and so we get d/dt#=2sine^[sin^2t]##cose^[sin^2t]#. Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1622 views around the world You can reuse this answer Creative Commons License