Find dy/dx y=e^(2x)sin(x)

Solution:

Step:1

Apply the differentiation operator to both sides of the given function: $\frac{d}{dx}(y) = \frac{d}{dx}(e^{2x}\sin(x))$.

Step:2

The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$.

Step:3

Proceed to differentiate the expression on the right-hand side.

Step:3.1

Utilize the Product Rule for differentiation: $\frac{d}{dx}[f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}[f(x)]$, where $f(x) = e^{2x}$ and $g(x) = \sin(x)$. Thus, we have $e^{2x}\frac{d}{dx}[\sin(x)] + \sin(x)\frac{d}{dx}[e^{2x}]$.

Step:3.2

The derivative of $\sin(x)$ with respect to $x$ is $\cos(x)$. This gives us $e^{2x}\cos(x) + \sin(x)\frac{d}{dx}[e^{2x}]$.

Step:3.3

Apply the Chain Rule for differentiation, which states $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$, where $f(x) = e^x$ and $g(x) = 2x$.

Step:3.3.1

Introduce a substitution, let $u = 2x$. The expression becomes $e^{2x}\cos(x) + \sin(x)(\frac{d}{du}[e^u]\frac{d}{dx}[2x])$.

Step:3.3.2

Differentiate using the Exponential Rule: $\frac{d}{du}[e^u] = e^u\ln(e)$, since $a = e$. This simplifies to $e^{2x}\cos(x) + \sin(x)(e^u\frac{d}{dx}[2x])$.

Step:3.3.3

Substitute back $u$ with $2x$ to get $e^{2x}\cos(x) + \sin(x)(e^{2x}\frac{d}{dx}[2x])$.

Step:3.4

Perform the differentiation.

Step:3.4.1

As $2$ is a constant, the derivative of $2x$ with respect to $x$ is $2\frac{d}{dx}[x]$. This gives us $e^{2x}\cos(x) + \sin(x)(e^{2x}(2\frac{d}{dx}[x]))$.

Step:3.4.2

Apply the Power Rule: $\frac{d}{dx}[x^n] = nx^{n-1}$, where $n = 1$. This simplifies to $e^{2x}\cos(x) + \sin(x)(e^{2x}(2\cdot1))$.

Step:3.4.3

Simplify the expression.

Step:3.4.3.1

Multiply $2$ by $1$ to get $e^{2x}\cos(x) + \sin(x)(e^{2x}\cdot2)$.

Step:3.4.3.2

Rearrange to place the constant $2$ in front of $e^{2x}$, resulting in $e^{2x}\cos(x) + \sin(x)(2\cdot e^{2x})$.

Step:3.4.3.3

Combine like terms to obtain $e^{2x}\cos(x) + 2e^{2x}\sin(x)$.

Step:4

Express the derivative of $y$ as equal to the simplified right-hand side: $\frac{dy}{dx} = e^{2x}\cos(x) + 2e^{2x}\sin(x)$.

Step:5

Replace $\frac{dy}{dx}$ with $y'$ to denote the derivative of $y$ with respect to $x$: $y' = e^{2x}\cos(x) + 2e^{2x}\sin(x)$.

Knowledge Notes:

1. Product Rule: When differentiating the product of two functions, the derivative is the first function times the derivative of the second function plus the second function times the derivative of the first function.

2. Chain Rule: Used to differentiate composite functions. If a variable is a function of another variable (e.g., $u = g(x)$), then the derivative of a function of that variable (e.g., $f(u)$) is the derivative of the outer function with respect to the inner function times the derivative of the inner function with respect to $x$.

3. Exponential Rule: The derivative of $e^u$ with respect to $u$ is $e^u$ itself, and if $u$ is a function of $x$, then the derivative involves the chain rule.

4. Power Rule: When differentiating $x^n$ with respect to $x$, the result is $nx^{n-1}$.

5. Simplification: In calculus, it is often necessary to simplify expressions by combining like terms or rearranging terms to make the differentiation process clearer.

6. Notation: The notation $\frac{dy}{dx}$ and $y'$ are both used to denote the derivative of $y$ with respect to $x$.