What is the first (e^{t^2})^' ?

The first (e^{t^2})^' is e^{t^2}*2t

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(e^{t^2})^'

(e^{t^{2}})^{^{'}}

e^{t^{2}} \cdot 2 t

Solution steps

(e^{t^{2}})^{'}

Apply the chain rule:

e^{t^{2}} (t^{2})^{'}

= e^{t^{2}} (t^{2})^{'}

(t^{2})^{'} = 2 t

= e^{t^{2}} \cdot 2 t

y = \frac{2}{x - 8}, (10, 1)

\int \frac{2 x}{5 - x ^{4}} d x

y = 2 x - x^{2}, (2, 0)

\int_{- 1}^{1} x^{- 3} d x

\frac{\partial}{\partial x} (\frac{e ^{x} - e ^{y}}{e ^{x y}})