# What is the derivative of #f(t) = ((lnt)^2-t, t^2cost ) #?

##### 1 Answer

Nov 13, 2016

# f'(t) = (2t^2cost - t^3sint) / (2lnt - t ) #

#### Explanation:

Define

# {(x(t) = ln^2t-t), (y(t) = t^2cost) :} #

Using the chain rule and product rule we can differentiate

# dx/dt = 2(lnt)1/t - 1 #

# :. dx/dt = 2/tlnt - 1 #

And,

# dy/dt = (t^2)(-sint) + (2t)(cost) #

# :. dy/dt = 2tcost - t^2sint #

And Also, By the chain rule:

# dy/dx = dy/dt * dt/dx => (dy/dt) / (dx/dt) #

# :. dy/dx = (2tcost - t^2sint) / (2/tlnt - 1 ) #

# :. dy/dx = (2tcost - t^2sint) / (1/t(2lnt - t )) #

# :. dy/dx = (2t^2cost - t^3sint) / (2lnt - t ) #