Find the Derivative - d/d@VAR f(t)=e^(5t)sin(t)

Solution:

Step 1: Apply the Product Rule

Utilize the Product Rule for differentiation, which is given by $\frac{d}{dt}[f(t)g(t)] = f(t)\frac{d}{dt}[g(t)] + g(t)\frac{d}{dt}[f(t)]$, where $f(t) = e^{5t}$ and $g(t) = \sin(t)$. Thus, we have $e^{5t}\frac{d}{dt}[\sin(t)] + \sin(t)\frac{d}{dt}[e^{5t}]$.

Step 2: Derivative of $\sin(t)$

The derivative of $\sin(t)$ with respect to $t$ is $\cos(t)$. This yields $e^{5t}\cos(t) + \sin(t)\frac{d}{dt}[e^{5t}]$.

Step 3: Chain Rule for Exponential Function

Invoke the Chain Rule for differentiation, which can be expressed as $\frac{d}{dt}[f(g(t))] = f'(g(t))g'(t)$, where $f(t) = e^t$ and $g(t) = 5t$.

Step 3.1: Introduce a Substitution

Let $u = 5t$. Our expression becomes $e^{5t}\cos(t) + \sin(t)\left(\frac{d}{du}[e^u]\frac{d}{dt}[5t]\right)$.

Step 3.2: Exponential Rule

Apply the Exponential Rule, which states that the derivative of $a^u$ with respect to $u$ is $a^u\ln(a)$, where $a = e$. This simplifies to $e^{5t}\cos(t) + \sin(t)(e^u\frac{d}{dt}[5t])$.

Step 3.3: Replace $u$ with $5t$

Substitute back $u$ with $5t$ to get $e^{5t}\cos(t) + \sin(t)(e^{5t}\frac{d}{dt}[5t])$.

Step 4: Differentiate $5t$

Now, differentiate $5t$ with respect to $t$.

Step 4.1: Constant Multiple Rule

Since $5$ is a constant, the derivative of $5t$ with respect to $t$ is $5\frac{d}{dt}[t]$. We now have $e^{5t}\cos(t) + \sin(t)(e^{5t}(5\frac{d}{dt}[t]))$.

Step 4.2: Power Rule

Utilize the Power Rule, which says that the derivative of $t^n$ with respect to $t$ is $nt^{n-1}$, where $n=1$. This simplifies to $e^{5t}\cos(t) + \sin(t)(e^{5t}(5\cdot1))$.

Step 4.3: Simplify the Expression

Combine and simplify the terms.

Step 4.3.1: Multiply Constants

Multiply $5$ by $1$ to get $e^{5t}\cos(t) + \sin(t)(e^{5t}\cdot5)$.

Step 4.3.2: Rearrange Terms

Rearrange the terms to place $5$ before $e^{5t}$, resulting in $e^{5t}\cos(t) + \sin(t)(5\cdot e^{5t})$.

Step 4.3.3: Combine Like Terms

Finally, combine the terms to obtain the derivative: $e^{5t}\cos(t) + 5e^{5t}\sin(t)$.

Knowledge Notes:

The problem involves finding the derivative of a function that is the product of two other functions, $e^{5t}$ and $\sin(t)$. To solve this, we use several rules of differentiation:

1. Product Rule: This rule is used when differentiating a product of two functions. It states that the derivative of $f(t)g(t)$ is $f(t)g'(t) + g(t)f'(t)$.

2. Chain Rule: The chain rule is applied when differentiating a composite function. It states that the derivative of $f(g(t))$ is $f'(g(t))g'(t)$.

3. Exponential Rule: This rule is used when differentiating exponential functions. For the base $e$, the derivative of $e^u$ with respect to $u$ is $e^u$ itself.

4. Constant Multiple Rule: When differentiating a constant multiplied by a function, the constant can be factored out and the derivative of the function is taken normally.

5. Power Rule: This rule is used to differentiate functions of the form $t^n$. The derivative is $nt^{n-1}$.

In this problem, we applied these rules in sequence to find the derivative of the given function. The solution steps reflect the systematic application of these rules, along with algebraic simplification to arrive at the final result.