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

Question

Scarlett Allison · Accepted Answer

# (dY)/(dT) = (e^(-T))/sqrt(1-e^(-2T)) #

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: # (dY)/(dT) = d/(dT) cos^(-1)(e^(-T)) # # \ \ \ \ \ \ = d/(dT) arccos(e^(-T))# # \ \ \ \ \ \ = -1/sqrt(1-(e^(-T))^2) * d/(dT)( e^(-T)) # # \ \ \ \ \ \ = -(-e^(-T))/sqrt(1-(e^(-T))^2) # # \ \ \ \ \ \ = (e^(-T))/sqrt(1-e^(-2T)) #

Isaiah Allen · Answer

undefined To find the derivative of $ Y = \cos^{-1}(e^{-T}) $, we'll use the chain rule.

Let $ u = e^{-T} $. Then, $ \frac{du}{dT} = -e^{-T} $.

Now, applying the chain rule, we have:

$$ \frac{d}{dT}(\cos^{-1}(u)) = \frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dT} $$

Substituting $ u = e^{-T} $ and $ \frac{du}{dT} = -e^{-T} $, we get:

$$ \frac{d}{dT}(\cos^{-1}(e^{-T})) = \frac{-1}{\sqrt{1 - (e^{-T})^2}} \cdot (-e^{-T}) $$

Simplify:

$$ \frac{d}{dT}(\cos^{-1}(e^{-T})) = \frac{e^{-T}}{\sqrt{1 - e^{-2T}}} $$

Isabella Ackerman · Answer

undefined To find the derivative of $ Y = \cos^{-1}(e^{-T}) $, we can use the chain rule.

Let $ u = e^{-T} $, then $ \frac{du}{dT} = -e^{-T} $.

Now, applying the chain rule:
$$ \frac{d}{dT}(\cos^{-1}(u)) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dT} $$

Substitute $ u = e^{-T} $ into the equation:
$$ \frac{d}{dT}(\cos^{-1}(e^{-T})) = -\frac{1}{\sqrt{1-(e^{-T})^2}} \cdot (-e^{-T}) $$

Simplify:
$$ \frac{d}{dT}(\cos^{-1}(e^{-T})) = \frac{e^{-T}}{\sqrt{1-e^{-2T}}} $$

Therefore, the derivative of $ Y = \cos^{-1}(e^{-T}) $ with respect to $ T $ is $ \frac{e^{-T}}{\sqrt{1-e^{-2T}}} $.