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What is the derivative of #f(t) = (t^3-e^(1-t) , tan^2t ) #?

Answer 1

Emilia Aguilar

#(dy)/(dx)=(3e^2+e^(1-t))/(2tant*sec^2t)#.

#f(t)=(t^3-e^(1-t),tan^2t)# is a parametric function in which both #x# and #y# are function of #t#.

As such #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#.

Here #(dy)/(dt)=3e^2-(-1)*e^(1-t)=3e^2+e^(1-t)# and

#(dx)/(dt)=2tant*sec^2t#

Hence #(dy)/(dx)=(3e^2+e^(1-t))/(2tant*sec^2t)#.

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Answer 2

Charles Anctil

To find the derivative of f(t) = (t^3 - e^(1-t), tan^2(t)):

Differentiate each component of the function separately with respect to t.
For the first component, differentiate t^3 - e^(1-t) with respect to t using the power rule and the chain rule.
For the second component, differentiate tan^2(t) with respect to t using the chain rule and the power rule.
Express the derivatives as a pair of functions in terms of t.

The derivative of f(t) = (t^3 - e^(1-t), tan^2(t)) is: f'(t) = (3t^2 + e^(1-t), 2tan(t)sec^2(t))

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