Question

What is the differentiation of ${{e}^{a{{x}^{2}}+b}}$ ?

Answer 1

We have been asked in the problem to find the differential of an exponential function. We will apply the chain rule of differentiation in our problem, assuming ‘a’ and ‘b’ are constants. The chain rule of differentiation can be given as: \dfrac{d\left[ f\left\\{ g\left( x \right) \right\\} \right]}{dx}=\dfrac{d\left[ f\left\\{ g\left( x \right) \right\\} \right]}{d\left[ g\left( x \right) \right]}\times \dfrac{d\left[ g\left( x \right) \right]}{dx} . We will also make use of the fact that the differentiation of any constant is equal to zero.

Complete step-by-step answer:
The expression given to us in our problem is equal to: ${{e}^{a{{x}^{2}}+b}}$ . Now, let us first assign some terms that we are going to use later in our problem.
Let the given term on which we need to operate a differential be given by ‘y’ . Here, ‘y’ is given to us as:
$\Rightarrow y={{e}^{a{{x}^{2}}+b}}$

Then, we need to find the differential of ‘y’ with respect to ‘x’. This can be done as follows:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{e}^{a{{x}^{2}}+b}} \right]}{dx}$

Applying the chain rule of differentiation on the second term, our expression can be written as:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{e}^{a{{x}^{2}}+b}} \right]}{d\left( a{{x}^{2}}+b \right)}\times \dfrac{d\left( a{{x}^{2}}+b \right)}{dx}$

Using the formula for differential of an exponential function, that is equal to:
$\Rightarrow \dfrac{d{{e}^{\theta }}}{d\theta }={{e}^{\theta }}$
We get our equation as:
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{e}^{a{{x}^{2}}+b}} \right]}{d\left( a{{x}^{2}}+b \right)}\times \dfrac{d\left( a{{x}^{2}}+b \right)}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\left( {{e}^{a{{x}^{2}}+b}} \right)\times \dfrac{d\left( a{{x}^{2}}+b \right)}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\left( {{e}^{a{{x}^{2}}+b}} \right)\times \left[ \dfrac{d\left( a{{x}^{2}} \right)}{dx}+\dfrac{db}{dx} \right] \\\ \end{aligned}$

Now, using the formula for differentiating an expression raised to some power:
$\Rightarrow \dfrac{d{{\theta }^{n}}}{d\theta }=n{{\theta }^{n-1}}$

And, using the fact that the differentiation of any constant term is equal to zero. Our equation simplifies into:
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\left( {{e}^{a{{x}^{2}}+b}} \right)\times \left[ 2ax+0 \right] \\\ & \therefore \dfrac{dy}{dx}=2ax{{e}^{a{{x}^{2}}+b}} \\\ \end{aligned}$

Thus, our final expression comes out to be $2ax{{e}^{a{{x}^{2}}+b}}$.
Hence, the differential of the expression ${{e}^{a{{x}^{2}}+b}}$ comes out to be $2ax{{e}^{a{{x}^{2}}+b}}$.

Note: While applying the chain rule, we should always make sure we are differentiating our fundamental equation with respect to a correct function. Finding out this function is the key to our solution. Also, while applying different formulas in the same expression, we should make sure that the formulas and calculations are correct.