What is the derivative of #tan^2(sinx)#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer Luke Phillips Aug 8, 2017 #"d"/("d"x) tan^2(sin(x)) = 2tan(sin(x))sec^2(sin(x))cos(x)# Explanation: By the chain rule, #"d"/("d"x) tan^2(sin(x)) = 2*tan(sin(x))* "d"/("d"x) (tan(sin(x)))#, #"d"/("d"x) tan^2(sin(x)) = 2tan(sin(x))sec^2(sin(x))*"d"/("d"x)(sin(x))#, #"d"/("d"x) tan^2(sin(x)) = 2tan(sin(x))sec^2(sin(x))cos(x)#. Answer link Related questions What is the derivative of #y=cos(x)# ? What is the derivative of #y=tan(x)# ? How do you find the 108th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x)# from first principle? How do you find the derivative of #y=cos(x^2)# ? How do you find the derivative of #y=e^x cos(x)# ? How do you find the derivative of #y=x^cos(x)#? How do you find the second derivative of #y=cos(x^2)# ? How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x) Impact of this question 6220 views around the world You can reuse this answer Creative Commons License