# 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# ?

What is the derivative of

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Note that

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Given that

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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)} ).

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