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The chain rule is a vital technique in calculus for differentiating composite functions. It helps us differentiate when a function is nested inside another function. In our given problem, we have the expression \( w = \sin^4 t - \cos t \tan t \), where \( w \) depends indirectly on \( t \) through \( \sin^4 t \) and involves chain differentiation.Let's break down the process:

• Identify the outer function and the inner function.
• For \( \sin^4 t \), the outer function is \( u^4 \) and the inner function is \( u = \sin t \).
• First, differentiate the power, giving \( 4u^3 \), and then, using the chain rule, multiply it by the derivative of \( \sin t \) which is \( \cos t \).
• This gives us \( 4 \sin^3 t \cos t \) as the derivative of \( \sin^4 t \).

This straightforward use of the chain rule makes differentiating composite trigonometric expressions seamless. Remember to follow the rule: differentiate the outer, then multiply by the derivative of the inner.

Differentiating products of functions requires the product rule. This rule is used when you're dealing with two separate functions multiplied together within an expression. In our exercise, the term \( -\cos t \tan t \) requires the product rule to be differentiated correctly.Here's how it works:

• Consider \( u = \cos t \) and \( v = \tan t \).
• The product rule states that \( (uv)' = u'v + uv' \).
• Differentiating gives \( u' = -\sin t \) and \( v' = \sec^2 t \).
• Apply the product rule: \( -\sin t \tan t - \cos t \sec^2 t \).

Applying these derivatives to the expression ensures we factor in the way both functions change with respect to \( t \), maintaining accuracy. Differentiating each part methodically avoids mistakes and leads to a complete derivative.

Understanding trigonometric derivatives is crucial in solving calculus differentiation problems involving trigonometric functions. Trigonometric functions like \( \sin t \), \( \cos t \), and \( \tan t \) have specific derivative formulas, which are key to solving such calculus exercises.Key derivatives include:

• The derivative of \( \sin t \) is \( \cos t \).
• \( \cos t \) differentiates to \( -\sin t \).
• For \( \tan t \), its derivative is \( \sec^2 t \).

In our problem, knowing these derivatives helps in calculating \( \frac{dw}{dt} \) accurately. Incorporating these derivatives in conjunction with chain and product rules allows us to navigate and solve complicated trigonometric expressions. This ensures we grasp the changes in each trigonometric component effectively while differentiating. Make sure to familiarize yourself with these basic derivatives, as they are foundational tools in many calculus problems involving trigonometry.