Question

Compute the derivative of the given function. $$g(t)=\sin \left(t^{5}+\frac{1}{t}\right)$$

Short answer

The derivative is \( g'(t) = \cos\left(t^5 + \frac{1}{t}\right) \cdot \left(5t^4 - \frac{1}{t^2}\right) \).

Step-by-step solution

Identify the Function Composition

The given function is \( g(t) = \sin\left(t^5 + \frac{1}{t}\right) \). This is a composition of functions, where the inner function is \( u(t) = t^5 + \frac{1}{t} \) and the outer function is \( f(u) = \sin(u) \). We will use the chain rule to find the derivative of \( g(t) \).

Differentiate the Inner Function

Differentiate \( u(t) = t^5 + \frac{1}{t} \) with respect to \( t \). The derivative of \( t^5 \) is \( 5t^4 \) and the derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \). Thus, \( u'(t) = 5t^4 - \frac{1}{t^2} \).

Differentiate the Outer Function

Differentiate \( f(u) = \sin(u) \) with respect to \( u \). The derivative of \( \sin(u) \) is \( \cos(u) \). Thus, \( f'(u) = \cos(u) \).

Apply the Chain Rule

The chain rule states that the derivative of the composite function \( g(t) = f(u(t)) \) is \( g'(t) = f'(u(t)) \cdot u'(t) \). So, \( g'(t) = \cos(u(t)) \cdot (5t^4 - \frac{1}{t^2}) \). Substitute back \( u(t) = t^5 + \frac{1}{t} \) to get: \[ g'(t) = \cos\left(t^5 + \frac{1}{t}\right) \cdot \left(5t^4 - \frac{1}{t^2}\right) \]

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. Imagine you have two functions, an outer function and an inner function. The chain rule helps us differentiate a mix of these without separating them. If you have a composite function like \( g(t) = f(u(t)) \), the chain rule tells us that the derivative is found by differentiating the outer function \( f \) with respect to the inner function \( u \), then multiplying by the derivative of the inner function with respect to \( t \). This process is mathematically represented as: \( g'(t) = f'(u(t)) \, u'(t) \).
In our example, the outer function is \( f(u) = \sin(u) \) and the inner function is \( u(t) = t^5 + \frac{1}{t} \). So, we use the chain rule to get the derivative \( g'(t) = \cos(u(t)) \cdot (5t^4 - \frac{1}{t^2}) \).

Composite Functions

Composite functions combine two or more functions, where one function is applied to the result of another. For example, with \( g(t) = f(u(t)) \), the function \( u(t) \) produces output that becomes the input for function \( f \). This linkage is what makes composite functions intricate but interesting.
In our problem, the composite nature is evident as we have \( g(t) = \sin(t^5 + \frac{1}{t}) \). Here, the term \( t^5 + \frac{1}{t} \) is first calculated and then fed into the sine function. Understanding composite functions is essential because it allows us to recognize how different functions communicate, which is crucial for applying the chain rule effectively.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are based on the ratios of the sides of right-angled triangles. They are foundational in understanding wave patterns, circles, and periodic phenomena.
In calculus, trigonometric functions like \( \sin(x) \), \( \cos(x) \), and their derivatives play a significant role. They often appear in situations where periodic motion or circular motion is involved, like in our example with the function \( g(t) = \sin(t^5 + \frac{1}{t}) \). Such understanding can aid in solving a wide array of problems beyond simple derivative calculations.

Derivative of Sine Function

The derivative of the sine function is vital in calculus. The sine function, \( \sin(x) \), describes oscillations which are essential in physics, engineering, and more. Its rate of change, or derivative, is \( \cos(x) \).
This behavior means that as the sine function changes, its slope at any point can be determined by the cosine function. When we differentiate \( \sin(u) \) with respect to \( u \), we get \( \cos(u) \), as seen in our exercise: the derivative of the outer function of our composite function is \( \cos(t^5 + \frac{1}{t}) \). This, multiplied by the derivative of the inner function, gives us the complete derivative of the composite function.