What is the derivative of #f(t) = (t/(t+1) , te^(2t-1) ) #?

Question

What is the derivative of #f(t) = (t/(t+1)  , te^(2t-1) )   #?

Ezra Adams · Accepted Answer

Hence derivative of #f(t)=(t/(t+1),te^(2t-1))# is #(t+1)^2(1+2t)e^(2t-1)# In parametric form of function where #f(t)=(g(t),h(t))# #(dy)/(dx)=((dh)/(dt))/((dg)/(dt))# as such derivative of given function #f(t)=(t/(t+1),te^(2t-1))# is #(d/dt(te^(2t-1)))/(d/(dt)(t/(t+1))# #d/dt(te^(2t-1))=1xxe^(2t-1)+2txxe^(2t-1)=e^(2t-1)(1+2t)# and #d/(dt)(t/(t+1))=d/(dt)(1-1/(t+1))=1/(t+1)^2# Hence derivative of #f(t)=(t/(t+1),te^(2t-1))# is #(e^(2t-1)(1+2t))/(1/(t+1)^2)=(t+1)^2(1+2t)e^(2t-1)#

Emilia Aguilar · Answer

undefined To find the derivative of the vector-valued function $ \mathbf{f}(t) = \left( \frac{t}{t+1}, \, t e^{2t-1} \right) $, we differentiate each component function with respect to $ t $.

1. Differentiate the first component function $ \frac{t}{t+1} $ with respect to $ t $:
$$ \frac{d}{dt}\left( \frac{t}{t+1} \right) = \frac{(t+1)(1) - t(1)}{(t+1)^2} = \frac{1}{(t+1)^2} $$

2. Differentiate the second component function $ t e^{2t-1} $ with respect to $ t $ using the product rule:
$$ \frac{d}{dt}\left( t e^{2t-1} \right) = e^{2t-1} + t \cdot \frac{d}{dt}(e^{2t-1}) $$
Using the chain rule $ \frac{d}{dt}(e^{2t-1}) = e^{2t-1} \cdot \frac{d}{dt}(2t-1) = 2e^{2t-1} $:
$$ \frac{d}{dt}\left( t e^{2t-1} \right) = e^{2t-1} + 2te^{2t-1} $$

Therefore, the derivative of $ \mathbf{f}(t) $ is:
$$ \mathbf{f}'(t) = \left( \frac{1}{(t+1)^2}, \, e^{2t-1} + 2te^{2t-1} \right) $$