Differentiate the following expression with respect to x:

[tex]$\frac{e^{x^2} \sqrt{\sin x}}{(2 x+1)^3}$[/tex]

Apply the quotient rule to the function f ( x ) = ( 2 x + 1 ) 3 e x 2 s i n x ​ ​ .
Find the derivative of the numerator u ( x ) = e x 2 sin x ​ using the product rule and chain rule.
Find the derivative of the denominator v ( x ) = ( 2 x + 1 ) 3 using the chain rule.
Substitute the derivatives into the quotient rule formula and simplify the expression, resulting in: ( 2 x + 1 ) 3 2 x e x 2 sin x ​ ​ + 2 ( 2 x + 1 ) 3 sin x ​ e x 2 cos x ​ − ( 2 x + 1 ) 4 6 e x 2 sin x ​ ​ ​

Explanation

Problem Setup We are asked to find the derivative of the function

f ( x ) = ( 2 x + 1 ) 3 e x 2 s i n x ​ ​
with respect to x .

Applying the Quotient Rule To differentiate this function, we will use the quotient rule, which states that if f ( x ) = v ( x ) u ( x ) ​ , then

f ′ ( x ) = v ( x ) 2 u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) ​
In our case, u ( x ) = e x 2 sin x ​ and v ( x ) = ( 2 x + 1 ) 3 .

Finding u'(x) First, we need to find the derivative of u ( x ) . We will use the product rule, which states that if u ( x ) = a ( x ) b ( x ) , then u ′ ( x ) = a ′ ( x ) b ( x ) + a ( x ) b ′ ( x ) . Here, a ( x ) = e x 2 and b ( x ) = sin x ​ .

So, u ′ ( x ) = ( e x 2 ) ′ sin x ​ + e x 2 ( sin x ​ ) ′ .

Derivatives of a(x) and b(x) Now, we find the derivatives of a ( x ) and b ( x ) .

( e x 2 ) ′ = e x 2 ( 2 x ) = 2 x e x 2 (using the chain rule)
( sin x ​ ) ′ = 2 s i n x ​ 1 ​ ( sin x ) ′ = 2 s i n x ​ c o s x ​ (using the chain rule)

Substituting Back Substituting these back into the expression for u ′ ( x ) , we get

u ′ ( x ) = 2 x e x 2 sin x ​ + e x 2 2 s i n x ​ c o s x ​

Finding v'(x) Next, we find the derivative of v ( x ) = ( 2 x + 1 ) 3 . Using the chain rule,

v ′ ( x ) = 3 ( 2 x + 1 ) 2 ( 2 ) = 6 ( 2 x + 1 ) 2

Applying the Quotient Rule Formula Now, we substitute u ( x ) , v ( x ) , u ′ ( x ) , and v ′ ( x ) into the quotient rule formula:

f ′ ( x ) = ( 2 x + 1 ) 6 ( 2 x e x 2 s i n x ​ + e x 2 2 s i n x ​ c o s x ​ ) ( 2 x + 1 ) 3 − e x 2 s i n x ​ ( 6 ( 2 x + 1 ) 2 ) ​

Simplifying the Expression Finally, we can simplify the expression:

f ′ ( x ) = ( 2 x + 1 ) 3 2 x e x 2 s i n x ​ ​ + 2 ( 2 x + 1 ) 3 s i n x ​ e x 2 c o s x ​ − ( 2 x + 1 ) 4 6 e x 2 s i n x ​ ​

Final Answer Therefore, the derivative of the given function is:

( 2 x + 1 ) 3 2 x e x 2 sin x ​ ​ + 2 ( 2 x + 1 ) 3 sin x ​ e x 2 cos x ​ − ( 2 x + 1 ) 4 6 e x 2 sin x ​ ​ ​
Examples
In physics, when analyzing the motion of a damped oscillator with a time-dependent forcing function, you might encounter functions similar to the one in this problem. Differentiating such functions helps determine the velocity and acceleration of the oscillator, which are crucial for understanding its dynamic behavior. For example, if the function represents the displacement of the oscillator, its derivative gives the velocity, and the second derivative gives the acceleration. This analysis is essential for designing and controlling mechanical and electrical systems.

Answered by GinnyAnswer | 2025-07-03

To find the derivative of f ( x ) = ( 2 x + 1 ) 3 e x 2 s i n x ​ ​ , we use the quotient rule. We differentiate the numerator and denominator using product and chain rules, and substitute into the quotient rule formula to arrive at the final derivative. The simplified result is f ′ ( x ) = ( 2 x + 1 ) 3 2 x e x 2 s i n x ​ ​ + 2 ( 2 x + 1 ) 3 s i n x ​ e x 2 c o s x ​ − ( 2 x + 1 ) 4 6 e x 2 s i n x ​ ​ .
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Answered by Anonymous | 2025-07-04