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

Question

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

Luke Alvarez · Accepted Answer

# 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!

Charles Arocha · Answer

#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)#

Aurora Alvarado · Answer

#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}#

Charles Anctil · Answer

undefined To find the derivative of $ y = \ln(t^2 + 4) - \frac{1}{2} \arctan(\frac{t}{2}) $, use the chain rule and derivative formulas for natural logarithm and arctangent:

$ \frac{dy}{dt} = \frac{1}{t^2 + 4} \cdot 2t - \frac{1}{2} \cdot \frac{1}{1 + (\frac{t}{2})^2} \cdot \frac{1}{2} $

Simplify:

$ \frac{dy}{dt} = \frac{2t}{t^2 + 4} - \frac{1}{4 + \frac{t^2}{4}} $

Combine the fractions:

$ \frac{dy}{dt} = \frac{2t}{t^2 + 4} - \frac{1}{\frac{4t^2 + 16}{4}} $

$ \frac{dy}{dt} = \frac{2t}{t^2 + 4} - \frac{4}{4t^2 + 16} $

Combine terms with common denominator:

$ \frac{dy}{dt} = \frac{2t(4t^2 + 16) - 4(t^2 + 4)}{(t^2 + 4)(4t^2 + 16)} $

Simplify the numerator:

$ \frac{dy}{dt} = \frac{8t^3 + 32t - 4t^2 - 16}{(t^2 + 4)(4t^2 + 16)} $

Combine like terms:

$ \frac{dy}{dt} = \frac{8t^3 - 4t^2 + 32t - 16}{(t^2 + 4)(4t^2 + 16)} $

So, the derivative of $ y = \ln(t^2 + 4) - \frac{1}{2} \arctan(\frac{t}{2}) $ is $ \frac{8t^3 - 4t^2 + 32t - 16}{(t^2 + 4)(4t^2 + 16)} $.