# What is the derivative of Y = Cos^-1 (e^-T)?

Mar 22, 2018

$\frac{\mathrm{dY}}{\mathrm{dT}} = \frac{{e}^{- T}}{\sqrt{1 - {e}^{- 2 T}}}$

#### Explanation:

We use the following derivatives:

 {: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (af(x), af'(x),a " constant"), (e^(ax), ae^(ax), a " constant)"), (cos^(-1)x, -1/sqrt(1-x^2), ), (f(g(x)), f'(g(x)) \ g'(x),"(Chain rule)" ) :}

So that:

$\frac{\mathrm{dY}}{\mathrm{dT}} = \frac{d}{\mathrm{dT}} {\cos}^{- 1} \left({e}^{- T}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{d}{\mathrm{dT}} \arccos \left({e}^{- T}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{1}{\sqrt{1 - {\left({e}^{- T}\right)}^{2}}} \cdot \frac{d}{\mathrm{dT}} \left({e}^{- T}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{- {e}^{- T}}{\sqrt{1 - {\left({e}^{- T}\right)}^{2}}}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{e}^{- T}}{\sqrt{1 - {e}^{- 2 T}}}$