Question

Find \[\dfrac{d\left( {{x}^{2}}{{e}^{x}}\sin x \right)}{dx}\] is
1. \[x{{e}^{x}}\left( 2\sin x+x\sin x+x\cos x \right)\]
2. \[x{{e}^{x}}\left( 2\sin x+x\sin x-\cos x \right)\]
3. \[x{{e}^{x}}\left( 2\sin x+x\sin x+\cos x \right)\]
4. None of these

Answer 1

Hint: To find the derivation of the given function (expression) , we can solve it by applying product rule for the three functions which we know can be solve by the formula that
\[\dfrac{d\left( uvw \right)}{dx}=u'vw+uv'w+uvw'\]
The three functions are \[{{x}^{2}}\], \[{{e}^{x}}\] and \[\sin x\] ,and to find the derivative we do it by substituting the functions and finding its derivative and applying those values in the formula. When you substitute the values in the formula in that way we can use the standard derivative and chain rule and find the answer needed.

Complete step-by-step answer:
In this question we need to differentiate \[{{x}^{2}}{{e}^{x}}\sin x\]. We can write this function and its derivative mathematically as,
\[\dfrac{d\left( {{x}^{2}}{{e}^{x}}\sin x \right)}{dx}\]
Now we know that we need to differentiate the above given expression. We apply the product rule of differentiation to find the solution. Now we know the formula that \[\dfrac{d\left( uvw \right)}{dx}=u'vw+uv'w+uvw'\] therefore we know that we can write the derivative of this given function as
\[\dfrac{d\left( {{x}^{2}}{{e}^{x}}\sin x \right)}{dx}={{e}^{x}}\sin x\dfrac{d\left( {{x}^{2}} \right)}{dx}+{{x}^{2}}\sin x\dfrac{d\left( {{e}^{x}} \right)}{dx}+{{x}^{2}}{{e}^{x}}\dfrac{d\left( \sin x \right)}{dx}\]
Let us start by finding the derivatives needed of the above functions individually. Let us consider
\[{{e}^{x}}\sin x\dfrac{d\left( {{x}^{2}} \right)}{dx}\]
Now for this we know that derivative of \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\] . Therefore we can write this above given function as
\[{{e}^{x}}\sin x\dfrac{d\left( {{x}^{2}} \right)}{dx}={{e}^{x}}\sin x\times 2x\]
Now after this lets consider the next function that is
\[{{x}^{2}}\sin x\dfrac{d\left( {{e}^{x}} \right)}{dx}\]
Now for this we know that the derivative of \[\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}\]. Therefore using this formula to find the derivative we get that
\[{{x}^{2}}\sin x\dfrac{d\left( {{e}^{x}} \right)}{dx}={{x}^{2}}{{e}^{x}}\sin x\]
Now lastly the last function whose derivative we need to find is \[{{x}^{2}}{{e}^{x}}\dfrac{d\left( \sin x \right)}{dx}\] and we know that derivative of sine is cosine so we get that
\[{{x}^{2}}{{e}^{x}}\dfrac{d\left( \sin x \right)}{dx}={{x}^{2}}{{e}^{x}}\sin x\]
We next add the three derivative to get the answer needed which is
\[\dfrac{d\left( {{x}^{2}}{{e}^{x}}\sin x \right)}{dx}=2x{{e}^{x}}\sin x+{{x}^{2}}{{e}^{x}}\sin x+{{x}^{2}}{{e}^{x}}\cos x\]
Therefore
 \[\dfrac{d\left( {{x}^{2}}{{e}^{x}}\sin x \right)}{dx}\] = \[x{{e}^{x}}\left( 2\sin x+x\sin x+x\cos x \right)\]
Hence this is the answer to this question i.e. this is the derivative of the function.
So, the correct answer is “Option 1”.

Note: Students must know the formulas and properties of basic functions to be easily able to solve these sums. Therefore use standard formulas known to solve these questions. We can also solve this question alternatively using the basic product rule. The formula we used here is derived from the basic formula of product rule of differentiation.