Question

How do you find the derivative of \[g\left( t \right)={{e}^{\dfrac{-3}{{{t}^{2}}}}}\]?

Answer 1

Hint: Assume the exponent of e as f (t) and write the given function g (t) in the form: \[g\left( t \right)={{e}^{f\left( t \right)}}\]. Now, differentiate both the sides with respect to the variable t and use the chain rule of differentiation given as: $\dfrac{d\left( {{e}^{f\left( t \right)}} \right)}{dt}={{e}^{f\left( t \right)}}\times f'\left( t \right)$ to find the derivative of g (t). Here, f’(t) is the derivative of the assumed function f (t).

Complete step by step solution:
Here, we have been provided with the function \[g\left( t \right)={{e}^{\dfrac{-3}{{{t}^{2}}}}}\] and we are asked to differentiate it. Here we are going to use the chain rule of differentiation to get the answer.
\[\because g\left( t \right)={{e}^{\dfrac{-3}{{{t}^{2}}}}}\]
Clearly, we can see that we have g (t) as a function of variable t. Now, we have another function as the exponent of e, so let us assume the exponent of e as f (t). So, we have,
\[\Rightarrow g\left( t \right)={{e}^{f\left( t \right)}}\], here \[f\left( t \right)=\dfrac{-3}{{{t}^{2}}}\].
Now, we can assume the above function as a composite function, so we need to use the chain rule of differentiation to get the required derivative. Therefore, using the formula: $\dfrac{d\left( {{e}^{f\left( t \right)}} \right)}{dt}=\dfrac{d\left( {{e}^{f\left( t \right)}} \right)}{d\left( f\left( t \right) \right)}\times f'\left( t \right)$, we have on differentiating both sides with respect to t,
\[\begin{align}
  & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}=\dfrac{d\left( {{e}^{f\left( t \right)}} \right)}{d\left[ f\left( t \right) \right]}\times f'\left( t \right) \\
 & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}=\dfrac{d\left( {{e}^{f\left( t \right)}} \right)}{d\left[ f\left( t \right) \right]}\times \dfrac{d\left[ f\left( t \right) \right]}{dt} \\
\end{align}\]
Substituting back the assumed value of f (t) and using the formula: $\dfrac{d\left( {{e}^{f\left( t \right)}} \right)}{dt}={{e}^{f\left( t \right)}}\times f'\left( t \right)$, we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}=\dfrac{d\left( {{e}^{\dfrac{-3}{{{t}^{2}}}}} \right)}{d\left[ \dfrac{-3}{{{t}^{2}}} \right]}\times \dfrac{d\left[ \dfrac{-3}{{{t}^{2}}} \right]}{dt} \\
 & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}={{e}^{\dfrac{-3}{{{t}^{2}}}}}\times \dfrac{d\left[ -3{{t}^{-2}} \right]}{dt} \\
\end{align}\]
Since -3 is a constant, it can be taken out of the derivative. Using the formula: $\dfrac{d\left( {{t}^{n}} \right)}{dt}=n{{t}^{n-1}}$, we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}={{e}^{\dfrac{-3}{{{t}^{2}}}}}\times \left( -3 \right)\left( -2 \right){{t}^{-2-1}} \\
 & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}={{e}^{\dfrac{-3}{{{t}^{2}}}}}\times 6{{t}^{-3}} \\
 & \Rightarrow \dfrac{d\left( g\left( t \right) \right)}{dt}=\dfrac{6{{e}^{\dfrac{-3}{{{t}^{2}}}}}}{{{t}^{3}}} \\
\end{align}\]
Hence, the above relation is required answer

Note: One may note that the general formula for the derivative of an exponential function ${{a}^{t}}$ is given as \[\dfrac{d\left[ {{a}^{t}} \right]}{dt}={{a}^{t}}\operatorname{lna}\], where ‘a’ is any real number other than 0. The exponential function ${{e}^{t}}$ is a special case of the given situation where we have a = e = 2.71. In this case we use the formula $\operatorname{lne}=1$ to simplify. Remember that in the formula $\dfrac{d\left( {{t}^{n}} \right)}{dt}=n{{t}^{n-1}}$ ‘n’ must not be 0. If that happens then the value of ${{t}^{n}}$ will become 1 which is a constant and we know that the derivative of a constant term is 0.