質問: How do you differentiate # f(x)=tan(e^((lnx-2)^2 ))# using the chain rule.? ===========================================================================
回答: #((2sec^2(e^((ln(x)-2)^2) )e^((ln(x)-2)^2)(lnx-2))/x)# #### Explanation: #d/dx (tan(e^((ln(x)-2)^2)))=sec^2(e^((ln(x)-2)^2))*d/dx((e^((ln(x)-2)^2))# =#sec^2(e^((ln(x)-2)^2))e^(((ln(x)-2))^2 )*d/dx(ln(x)-2)^2# =#sec^2(e^((ln(x)-2)^2))e^(((ln(x)-2))^2)2(lnx-2)* d/dx(lnx-2)# =#(sec^2(e^((ln(x)-2)^2))e^(((ln(x)-2))^2)2(lnx-2) *1/x)# =#((2sec^2(e^((ln(x)-2)^2) )e^((ln(x)-2)^2)(lnx-2))/x)#