微積分導數

求y=Cotx在x=π/4之切線方程式
Sol
y=Cotx
y’=－Csc^2 x
y-Cot(π/4)=－Csc^2 (π/4)(x－π/4)
y－1=－2(x－π/4)

求d/dt{Cos^2【Cos(Cost)】}
Sol
dCos^2【Cos(Cost)】/dt
=｛dCos^2【Cos(Cost)】/dCos【Cos(Cost)】｝* ｛dCos【Cos(Cost)】/d【Cos(Cost)】｝
* dCos(Cost)/ dCost*dCost/dt
=2Cos【Cos(Cost)】*｛－Sin【Cos(Cost)】｝*[－Sin(Cost)]*(－Sint)
=－2Cos【Cos(Cost)】*Sin【Cos(Cost)】*Sin(Cost)*Sint
=－Cos｛Sin【2Cos(Cost)】｝*Sin(Cost)*Sint

斜率 y' = -csc^2 x = -1/sin^2 π/4 =-2
x=π/4 , y=cotπ/4 =1 =>切點(π/4 ,1)
=> 切線方程式 y-1 =-2(x-π/4)