What is the derivative of #y = ln(t^2 + 4) - 1/2 arctan(t/2)#?

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What is the derivative of #y = ln(t^2 + 4) - 1/2 arctan(t/2)# with respec to #t#?

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Answer 1 · 2018-07-19T19:15:30

# dy/dt=(2t-1)/(t^2+4)#.

#y=ln(t^2+4)-1/2arctan(t/2)#.

#dy/dt=d/dt{ln(t^2+4)-1/2arctan(t/2)}#,

#=d/dt{ln(t^2+4)}-1/2d/dt{arctan(t/2)}#,

#=1/(t^2+4)*d/dt(t^2+4)-1/2*1/{1+(t/2)^2}*d/dt{t/2}#,

#=1/(t^2+4)*(2t)-1/2*1/(1+t^2/4)*1/2#,

#=(2t)/(t^2+4)-1/4*4/(4+t^2)#.

# rArr dy/dt=(2t-1)/(t^2+4)#, as desired!

Answer 2 · 2018-07-19T19:16:00

#f'(t)=(2t-1)/(t^2+4)#

Note that

#(ln(x))'=1/x# and #(arctan(x))'=1/(1+x^2)#
and additionally we use the chain rule

#(f(g(x)))'=f'(g(x))*g'(x)#
so we get

#f'(t)=(2t)/(t^2+4)-1/2*1/(1+(t/2)^2)*1/2#

note that

#1/4*1/(1+t^2/4)=1/4*4/(4+t^2)=1/(4+t^2)#
so we get

#y'=(2t-1)/(t^2+4)#

Answer 3 · 2018-07-19T19:29:44

#dy/dt={2t-1}/{t^2+4}#

Given that

#y=\ln(t^2+4)-1/2 \tan^{-1}(t/2)#

differentiating above function w.r.t. #t# using chain rule as follows

#dy/dt=d/dt(ln(t^2+4)-1/2 \tan^{-1}(t/2))#

#=1/{t^2+4}d/dt(t^2+4)-1/2\frac{1}{1+(t/2)^2}d/dt(t/2)#

#=1/{t^2+4}(2t)-1/2\frac{4}{t^2+4}(1/2)#

#={2t}/{t^2+4}-\frac{1}{t^2+4}#

#={2t-1}/{t^2+4}#