Question

Differentiate the following functions. $$ u=\sin ^{3} x $$

Short answer

Answer: The derivative of the function sin^3(x) with respect to x is 3sin^2(x)cos(x).

Step-by-step solution

Identify the inner and outer functions

In the given function sin^3(x), the inner function is sin(x) and the outer function is raising to the power of 3.

Use the chain rule for differentiation

According to the chain rule, if we have a composite function u(v(x)), then its derivative is: $$ \frac{d}{dx}(u\big(v(x)\big)) = u'\big(v(x)\big)\cdot v'(x) $$ In our case, the outer function u(v) is v^3, so u'(v) = 3v^2. The inner function v(x) is sin(x), and its derivative v'(x) is cos(x).

Apply the chain rule

Now we can apply the chain rule to find the derivative: $$ \frac{d}{dx}(\sin^3(x)) = 3(\sin(x))^2\cdot\cos(x) $$ So, the derivative of the given function is: $$ \frac{d}{dx}(\sin^3(x)) = 3\sin^2(x)\cos(x) $$

Key concepts

Chain Rule

The chain rule is an essential tool in calculus when dealing with composite functions. It is particularly useful when a function is nested inside another. In simple terms, the chain rule helps us find the derivative of a composition of two or more functions.
For a function that can be expressed as a composition like \(f(g(x))\), where \(f\) is the outer function and \(g\) is the inner function, the chain rule states that the derivative is the product of the derivative of the outer function at the inner function times the derivative of the inner function itself:

• First, differentiate the outer function while keeping the inner function intact.
• Multiply this by the derivative of the inner function.

Applying this to our example, \(u = \sin^{3}(x)\), we treat the outer function as \(v^3\), where \(v = \sin(x)\). The differentiation follows:
\[\frac{d}{dx}(v^3) = 3v^2\times v'(x) \]
Here, the inner derivative \(v'(x) = \cos(x)\). The chain rule ensures we can manage complex functions by breaking them down into simpler, more manageable parts.

Derivative of Trigonometric Functions

Differentiating trigonometric functions is a key part of calculus and it helps solve a variety of real-world problems. The basic trigonometric functions include sine, cosine, and tangent, among others. Each has a specific rule for differentiation:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• The derivative of \(\tan(x)\) is \(\sec^2(x)\).

In our original problem, \(\sin(x)\) is the trigonometric function of interest. When we apply the derivative to \(\sin(x)\), we get \(\cos(x)\). This step is crucial when using the chain rule, as it allows us to find the inner derivative needed to solve for the entire function. Thus, for \(u = \sin^3(x)\), the inner function's derivative \(v'(x)\) is \(\cos(x)\), which then plugs into the chain rule formula.

Composite Functions

Composite functions are formed when one function is applied to the result of another function. They are written in the form \(f(g(x))\) or often described using \(u(v(x))\), where \(f\) or \(u\) is the outer function and \(g\) or \(v\) is the inner function.
Understanding composite functions is crucial because many real-life situations require such compositions. Solving them involves understanding both functions separately and using techniques like the chain rule to differentiate them effectively.
For example, in our exercise, \(u = \sin^3(x)\) means we have a function, specifically \(v = \sin(x)\), nested inside another function, \(v^3\). Computing their derivatives involves treating the outer operation first, then working inward.
Knowing how composite functions operate helps to unravel potentially complicated expressions into simpler components, making calculus problems more approachable and solvable.