Differentiate the following with respect to \[x\]: \[{e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\]

Hint: Here, we will find the differentiation of the given function with respect to the variable. We will find the derivative of all the functions separately. First, we will find the derivative of the exponential function and then the derivative of the exponential of the trigonometric function. Then we will simplify it further to get the required value.

Formula Used:
We will use the following formulas:
1. \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = x{e^x}\]
2. \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
3. \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
4. \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
5. \[2\sin x\cos x = \sin 2x\]

Complete step-by-step answer:
Let the given function be \[y\].
So, we get \[y = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\]
Differentiating with respect to \[x\], we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{3{{\sin }^2}x - 2{{\cos }^2}x}}} \right)\]
By using the derivative formula \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = x{e^x}\], we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}} \cdot \dfrac{d}{{dx}}\left( {3{{\sin }^2}x - 2{{\cos }^2}x} \right)\]
We know that \[{\sin ^2}x = {\left( {\sin x} \right)^2}\].
So, rewriting the above equation, we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}} \cdot \left[ {3\dfrac{d}{{dx}}{{\left( {\sin x} \right)}^2} - 2\dfrac{d}{{dx}}{{\left( {\cos x} \right)}^2}} \right]\]
By using the derivative formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}} \cdot \left[ {3 \cdot 2\sin x \cdot \dfrac{d}{{dx}}\left( {\sin x} \right) - 2 \cdot 2\cos x\dfrac{d}{{dx}}\left( {\cos x} \right)} \right]\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\left[ {6\sin x \cdot \dfrac{d}{{dx}}\left( {\sin x} \right) - 4\cos x\dfrac{d}{{dx}}\left( {\cos x} \right)} \right]\]
By using the derivative formula \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] and \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\] , we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\left[ {6\sin x \cdot \cos x - 4\cos x \cdot \left( { - \sin x} \right)} \right]\]
We know that the product of two negative integers is a positive integer. Therefore, we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\left[ {6\sin x \cdot \cos x + 4\sin x \cdot \cos x} \right]\]
By adding the terms, we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\left[ {10\sin x \cdot \cos x} \right]\]
By rewriting the terms in terms of the trigonometric identity, we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\left[ {5 \cdot 2\sin x\cos x} \right]\]
Using the trigonometric formula, \[2\sin x\cos x = \sin 2x\], we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\left[ {5\sin 2x} \right]\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 5\sin 2x \cdot {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\]
 \[ \Rightarrow y' = 5\sin 2x \cdot {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\]
Therefore, the derivative of \[{e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\] is \[5\sin 2x \cdot {e^{3{{\sin }^2}x - 2{{\cos }^2}x}}\].

Note: We should know that if a function has two functions, then both the functions have to be differentiated separately. Differentiation is a method of finding the derivative the function and finding the rate of change of function with respect to one. Here we have found out the derivative of exponential function. Exponential function is a constant which is raised to some power. Exponential function is the inverse of logarithmic function.