## Question

###### Find Dy/dx Help i have no idea how to use the chain rule here? y=ln(tan^-1(ex))

Find Dy/dx Help i have no idea how to use the chain rule here? y=ln(tan^-1(ex))

## Answers

#(dy)/(dx)=e/((1+e^2x^2)tan^-1(ex))# ## Explanation:

We know that,

#color(red)((1) d/(dX)(lnX)=1/X#

#color(red)((2) d/(dX)(tan^-1X)=1/(1+X^2# Here,

#y=ln(tan^-1(ex))# Take,

#y=lnuand u=tan^-1(ex)#

#:.(dy)/(du)=1/u and(du)/(dx)=1/(1+(ex)^2)d/(dx)(ex)#

#where,u=tan^-1(ex) and color(orange)(d/(dx)(ex)=e#

#i.e. (dy)/(du)=1/(tan^-1(ex))and(du)/(dx)=e/(1+e^2x^2)#

#"Using "color(blue)"Chain Rule:"#

#color(blue)((dy)/(dx)=(dy)/(du)xx(du)/(dx)#

#:.(dy)/(dx)=1/(tan^-1(ex))xxe/(1+e^2x^2)#

#=>(dy)/(dx)=e/((1+e^2x^2)tan^-1(ex))# Note:

If

#y=ln(tan^-1(e^x)). then ,take,v=e^x=>(dv)/(dx)=color(orange)(d/(dx)(e^x)=e^x# .So,the answer will be

#=>(dy)/(dx)=e^x/((1+(e^x)^2)tan^-1(e^x))#

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