Find dy/dx y=e^(-8x)sin(x)

Solution:

Step 1:

Take the derivative of both sides with respect to $x$: $\frac{d}{dx}(y)=\frac{d}{dx}(e^{-8x}\sin (x))$.

Step 2:

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

Step 3:

Compute the derivative of the right-hand side.

Step 3.1:

Apply the Product Rule: $\frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}$, where $f(x)=e^{-8x}$ and $g(x)=\sin (x)$. Thus, we have $e^{-8x}\frac{d}{dx}(\sin (x))+\sin (x)\frac{d}{dx}(e^{-8x})$.

Step 3.2:

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$. This gives us $e^{-8x}\cos (x)+\sin (x)\frac{d}{dx}(e^{-8x})$.

Step 3.3:

Employ the Chain Rule for differentiation: $\frac{d}{dx}(f(g(x)))=f^{\prime }(g(x))g^{\prime }(x)$, where $f(x)=e^x$ and $g(x)=-8x$.

Step 3.3.1:

Let $u=-8x$. The expression becomes $e^{-8x}\cos (x)+\sin (x)(\frac{d}{du}(e^u)\frac{dx}{du})$.

Step 3.3.2:

Differentiate using the Exponential Rule: $\frac{d}{du}(e^u)=e^u\ln (e)$, where $a=e$. We get $e^{-8x}\cos (x)+\sin (x)(e^u\frac{d}{dx}(-8x))$.

Step 3.3.3:

Substitute back $u=-8x$. The expression simplifies to $e^{-8x}\cos (x)+\sin (x)(e^{-8x}\frac{d}{dx}(-8x))$.

Step 3.4:

Perform the differentiation.

Step 3.4.1:

Since $-8$ is a constant, the derivative of $-8x$ with respect to $x$ is $-8\frac{d}{dx}(x)$. We now have $e^{-8x}\cos (x)+\sin (x)(e^{-8x}(-8\frac{d}{dx}(x)))$.

Step 3.4.2:

Apply the Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$, where $n=1$. This yields $e^{-8x}\cos (x)+\sin (x)(e^{-8x}(-8\cdot 1))$.

Step 3.4.3:

Simplify the expression.

Step 3.4.3.1:

Multiply $-8$ by $1$ to get $e^{-8x}\cos (x)+\sin (x)(e^{-8x}\cdot -8)$.

Step 3.4.3.2:

Rearrange to place $-8$ in front of $e^{-8x}$, resulting in $e^{-8x}\cos (x)+\sin (x)(-8\cdot e^{-8x})$.

Step 3.4.3.3:

Combine like terms to obtain the final derivative: $e^{-8x}\cos (x)-8e^{-8x}\sin (x)$.

Step 4:

Express the equation with the left side equal to the right side: $y=e^{-8x}\cos (x)-8e^{-8x}\sin (x)$.

Step 5:

Substitute $\frac{dy}{dx}$ for $y$: $\frac{dy}{dx}=e^{-8x}\cos (x)-8e^{-8x}\sin (x)$.

Knowledge Notes:

1. Product Rule: When differentiating a product of two functions, $f(x)g(x)$, the derivative is $f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.

2. Chain Rule: This rule is used when differentiating a composite function, $f(g(x))$. The derivative is found by multiplying the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function.

3. Exponential Rule: The derivative of $e^u$, where $u$ is a function of $x$, is $e^u\cdot \frac{du}{dx}$. If $u$ is a constant times $x$, then $\frac{du}{dx}$ is just that constant.

4. Power Rule: For any real number $n$, the derivative of $x^n$ with respect to $x$ is $n{x}^{n-1}$.

5. Simplification: In calculus, simplifying expressions often involves combining like terms, factoring, and canceling common factors to make the result more concise and easier to understand.

6. Notation: It's important to maintain consistent notation throughout the differentiation process. For example, using $\frac{dy}{dx}$ to represent the derivative of $y$ with respect to $x$.

By understanding these rules and applying them correctly, one can find the derivative of a wide variety of functions.