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The chain rule is a fundamental concept in calculus used to differentiate composite functions. It tells us how to differentiate a function that is the composition of two or more smaller functions. The chain rule is handy when dealing with a function that involves "an inner and outer function." For instance, if we have a function of the form \( y = f(g(x)) \), the chain rule allows us to differentiate this by multiplying the derivative of the outer function evaluated at the inner function, \( f'(g(x)) \), by the derivative of the inner function, \( g'(x) \). For exercise (a), \( y = \tan \sqrt{t} \), the inner function is \( u = \sqrt{t} \) and differentiating it gives \( \frac{du}{dt} = \frac{1}{2\sqrt{t}} \). The outer function is \( \tan u \) and its derivative is \( \sec^2 u \). Applying the chain rule here combines these derivatives: \( \frac{dy}{dt} = \sec^2 \sqrt{t} \times \frac{1}{2\sqrt{t}} \). This approach simplifies handling complex derivatives by breaking them into manageable parts.

The quotient rule is another critical tool in calculus, particularly useful for functions expressed as a ratio of two differentiable functions. For a function given by a ratio \( y = \frac{u(x)}{v(x)} \), the quotient rule states that the derivative \( \frac{dy}{dx} \) is given by \( \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2} \). This can be remembered using the formula: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \). In part (b) of the exercise \( y = \frac{1-v^2}{1+v^2} \), applying the quotient rule involves identifying \( u = 1-v^2 \) and \( v = 1+v^2 \). First, find the derivatives \( u' = -2v \) and \( v' = 2v \). Substitute these into the quotient rule formula to achieve the derivative \( dy = \frac{-4v}{(1+v^2)^2} dv \), efficiently addressing a complex ratio.

Differentiating trigonometric functions is a key aspect of calculus and includes functions like sine, cosine, and tangent. These functions have specific rules we must remember. The derivatives of basic trigonometric functions are as follows:

• \( \frac{d}{dx} [\sin x] = \cos x \)
• \( \frac{d}{dx} [\cos x] = -\sin x \)
• \( \frac{d}{dx} [\tan x] = \sec^2 x \)

The exercise highlights the derivative of \( \tan x \), following the rule that differentiation results in \( \sec^2 x \). For instance, in exercise (a), when finding the derivative of \( y = \tan \sqrt{t} \), the derivative is expressed as \( \sec^2 \sqrt{t} \), while the application of chain rule multiplies this by the derivative of the inner function. These rules provide the foundational knowledge required for solving more complex calculus problems featuring trigonometric elements.

Approaching calculus problems methodically is crucial for success. The problems faced in the exercise required applying both the chain rule and quotient rule for differentiation. Here’s how to tackle such problems:

• Identify the type of function: Determine if you have a product, quotient, or composite function.
• Choose the right rule: Use chain rule for nested functions, quotient rule for divisions, and product rule for multiplications.
• Break it down: If the function is complex, separate it into "smaller, simpler" parts.
• Calculate derivatives carefully: Pay attention to deriving both outer and inner functions where applicable.
• Simplify results: Once differentiated, simplify the expression to achieve the cleanest form of the derivative.

Successfully solving calculus problems involves recognizing the structure of the function you're dealing with and methodically applying the appropriate rules. This systematic approach breaks challenging problems into simpler tasks, making calculus manageable and achievable.