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

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

3 Answers
Jul 19, 2018

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

Explanation:

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!

Jul 19, 2018

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

Explanation:

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)

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

Explanation:

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}