Question

How do you differentiate $f(x) = {e^{ - 5{x^2}}}$?

Answer 1

We will write the given function as the composition of two functions. Now, we will use the chain rule for differentiation of two functions and thus have the answer.

Complete step by step solution:
We are given that we are required to differentiate $f(x) = {e^{ - 5{x^2}}}$.
We can write the given function as the composition of two functions, one of them will be exponential and the other will be a square function.
Now, let us assume that $g(x) = {e^x}$ and $h(x) = - 5{x^2}$.
Therefore, we can now write the given function $f(x) = {e^{ - 5{x^2}}}$ as:
$\Rightarrow f(x) = g(h(x))$
Now, applying the chain rule on the above function, we have:-
$\Rightarrow f'(x) = g'(h(x))h'(x)$
Instead of h (x), we will put it in its place.
Putting the functions as above, we have the following equation:
$\Rightarrow \dfrac{d}{{dx}}\left( {{e^{ - 5{x^2}}}} \right) = \dfrac{d}{{dt}}\left( {{e^t}} \right) \times \dfrac{d}{{dx}}\left( { - 5{x^2}} \right)$
Simplifying the right hand side of the above equation, we will then obtain the following equation:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{e^{ - 5{x^2}}}} \right) = {e^t} \times \left( { - 10x} \right)$
Putting t back in the form of x as we assumed, we have:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{e^{ - 5{x^2}}}} \right) = - 10x{e^{ - 5{x^2}}}$
Thus, we have the required answer.

Note: The chain rule for differentiation of two functions, which we used in the above solution whose definition is given as follows:-
If we have two functions $f(x)$ and $g(x)$ which are differentiable and continuous, then we have the following expression:
$\Rightarrow \dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Students must also note that we converted the function in terms of t, so that it becomes easier for us to differentiate.
The students must also note that we have made use of the following facts in the above solution as well:-
1. The derivative of exponential function is exponential function only without any change that is $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$.
2. The derivative of algebraic function is given by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$.
3. We can always take out the constant while differentiating any function (as we took out – 5 common from h (x)) like as follows: $\dfrac{d}{{dx}}\left( {cf(x)} \right) = c\dfrac{d}{{dx}}\left( {f(x)} \right)$.