Find derivatives of the given functions : x= ln (1+t^2) y= t- arctgt ?

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Answer 1 · 2018-04-23T09:12:27

#(dy)/(dx)=t/2#

When we have #f(t)=(x(t),y(t))#,

then #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

Here #x(t)=ln(1+t^2)# and #(dx)/(dt)=(2t)/(1+t^2)#

and #y(t)=t-arctant# and therefore #(dy)/(dt)=1-1/(1+t^2)=t^2/(1+t^2)#

and hence #(dy)/(dx)=t^2/(2t)=t/2#

Answer 2 · 2018-04-23T09:13:43

See details below

We know that if #f(x)=lnh(x)#, then #f´(x)=(h´(x))/(h(x))# and if #g(x)=arctanx#, then #g´(x)=1/(1+x^2)#

In our case, we have

#x(t)=ln(1+t^2)# here #h(x)=1+t^2#

#dx/dt=(2t)/(1+t^2)#

#y(t)=t-arctant#

#dy/dt=1-1/(1+t^2)#

Now if you look for #dy/dx=(1-1/(1+t^2))/(2t/(1+t^2))=t/2#