Question

Evaluate the following derivatives. $$\frac{d}{d t}\left((\sin t)^{\sqrt{t}}\right)$$

Short answer

Question: Find the derivative of the function \((\sin t)^{\sqrt{t}}\) with respect to \(t\). Answer: \(\frac{d}{dt}((\sin t)^{\sqrt{t}}) = (\sin t)^{\sqrt{t}} \cdot \left(\frac{1}{2\sqrt{t}} \cdot \ln(\sin t) + \sqrt{t} \cdot \cot t\right)\)

Step-by-step solution

Rewrite the function to make it easier to differentiate

Let's rewrite the given function as $$y=(\sin t)^{\sqrt{t}} = e^{\sqrt{t} \cdot \ln(\sin t)}$$ This step makes the differentiation process easier using the chain rule.

Differentiate the new function

Now, we want to find the derivative \(y'\) of \(y = e^{\sqrt{t} \cdot \ln(\sin t)}\) with respect to \(t\). Using the chain rule, we have: $$y' = e^{\sqrt{t} \cdot \ln(\sin t)} \cdot \frac{d}{dt}(\sqrt{t} \cdot \ln(\sin t))$$

Differentiate the exponent

Now we need to differentiate the exponent \(\sqrt{t} \cdot \ln(\sin t)\) with respect to \(t\). To do this, we will use the product rule, which states \((u\cdot v)' = u'\cdot v + u\cdot v'\) where \(u = \sqrt{t}\) and \(v = \ln(\sin t)\). First, let's find the derivatives of \(u\) and \(v\): $$u' = \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}}$$ $$v' = \frac{d}{dt}(\ln(\sin t)) = \frac{\cos t}{\sin t} = \cot t$$ Now, apply the product rule: $$\frac{d}{dt}(\sqrt{t} \cdot \ln(\sin t)) =\frac{1}{2\sqrt{t}} \cdot \ln(\sin t) + \sqrt{t} \cdot \cot t$$

Substitute the derivative of the exponent back into the expression

Finally, substitute the result from Step 3 back into our expression for \(y'\): $$y' = e^{\sqrt{t} \cdot \ln(\sin t)} \cdot \left(\frac{1}{2\sqrt{t}} \cdot \ln(\sin t) + \sqrt{t} \cdot \cot t\right)$$

Rewrite the expression in original form

Now, we will rewrite the expression back in terms of the original function. We have: $$\frac{d}{dt}((\sin t)^{\sqrt{t}}) = (\sin t)^{\sqrt{t}} \cdot \left(\frac{1}{2\sqrt{t}} \cdot \ln(\sin t) + \sqrt{t} \cdot \cot t\right)$$ Thus, the derivative of the given function is: $$\frac{d}{dt}((\sin t)^{\sqrt{t}}) = (\sin t)^{\sqrt{t}} \cdot \left(\frac{1}{2\sqrt{t}} \cdot \ln(\sin t) + \sqrt{t} \cdot \cot t\right)$$

Key concepts

Chain Rule

The chain rule is a vital tool in calculus used for finding the derivative of composite functions. It allows us to differentiate functions that are nested within each other.
Imagine you have a function inside another function, like a matryoshka doll with a doll inside a bigger doll. The outside function depends on the inside function, and to differentiate this, we use the chain rule.

• The rule states that if you have a function of a function, say \( y = f(g(t)) \), the derivative \( y' \) is given by \( f'(g(t)) \cdot g'(t) \).
• In practical terms, you first differentiate the outer function, keeping the inner function constant.
• Then, multiply this by the derivative of the inner function.

By practicing the chain rule, you'll find it increasingly easier to tackle composite functions. Remember, it's all about meticulously applying each step.
This rule was an essential part of the solution for differentiating \( y = e^{\sqrt{t} \cdot \ln(\sin t)} \) as it guides us through the process of differentiating exponential functions with variables in the exponent.

Product Rule

The product rule is the hero whenever we deal with multiplying two functions. It helps us differentiate the product of two functions and is crucial when functions are not separable.
The product rule formula is a straightforward yet powerful concept:

Consider two functions \( u(t) \) and \( v(t) \). The derivative \((u \cdot v)'\) is \( u' \cdot v + u \cdot v' \).

• This means you differentiate one function while leaving the other untouched, and then switch roles.
• Finally, you add together these two parts.

In the given problem, we needed the product rule to handle the exponent \( \sqrt{t} \cdot \ln(\sin t) \). Here, \( u = \sqrt{t} \) and \( v = \ln(\sin t) \).
By successfully applying the product rule, we found the derivative of this product, allowing correct substitution back into the original expression.
Understanding and practicing the product rule makes differentiation of products less daunting and more structured.

Exponential Functions

Exponential functions, typically in the form of \( a^x \), where \( a \) is a constant, play a significant role in calculus and real-world applications. Differentiating exponential functions with variables in the exponent often requires logarithmic manipulation, as shown in this exercise.

• A common strategy is to rewrite the function using natural logarithms, expressing \( a^x \) as \( e^{x \cdot \ln(a)} \). This simplifies the differentiation process.
• By transforming the original function \( (\sin t)^{\sqrt{t}} \) into \( e^{\sqrt{t} \cdot \ln(\sin t)} \), the problem becomes more manageable using the chain and product rules.

Exponential functions are powerful in modeling growth processes, decay, and capturing dynamic changes in various fields.

Mastering their differentiation opens the door to understanding behaviors and trends depicted by these mathematical expressions. In calculus, they frequently pop up, hence knowing how to handle them becomes an invaluable skill.