91Ó°ÊÓ

The chain rule is a fundamental technique in calculus for differentiating composite functions. When you have two or more functions nested together, the chain rule helps find the derivative of the composed function.

For example, if you have a function like \(e^y\) where \(y\) is also a function of \((x)\), you can't just differentiate \(e^y\) directly with respect to \(x\). You first need to differentiate \(e^y\) with respect to \(y\), and then differentiate \(y\) with respect to \(x\).

Here's the basic idea of the chain rule in steps:

• Differentiate the outer function while keeping the inner function unchanged.
• Multiply the result by the derivative of the inner function.

In our problem, using the chain rule on \(e^y\), gives \(e^y \, \frac{dy}{dx}\), because \(e^y\) differentiates to \(e^y\) with respect to \(y\), and then you multiply by \(rac{dy}{dx}\) since \(y\) is a function of \(x\).

The product rule is a technique used in calculus for differentiating products of two functions. When you have a term like \(e^y \, \cos x\), it requires the product rule to find its derivative as it involves two multiplicative components.

Here’s the product rule formula:

• Let \(u(x)\) and \(v(x)\) be functions of \(x\).
• Then the derivative of their product is: \((uv)' = u'v + uv'\).

In our case, set \(u = e^y \,\) and \(v = \, \cos x\). Differentiating \(u\) involves the chain rule since \(e^y\) depends on \(y\), so you get \(u' = e^y \, \frac{dy}{dx}\). \(v'\) is simply \(-\sin x\) since the derivative of \(\cos x\) is \(-\sin x\).

The product rule calculation gives \(e^y \, (-\sin x) + \cos x \, (e^y \, \frac{dy}{dx})\). This result becomes part of the larger derivative computation process for the given equation.

Derivative computation is at the core of solving problems involving rate of change, slopes, and curved graphs, including implicit differentiation. In the context of implicit differentiation, it refers to finding the derivative of equations expressed in implicit form.

In the original exercise, each side of the equation, \(e^y \, \cos x = 1 + \sin(xy)\), needs to be differentiated with respect to \(x\).

Using a combination of the chain rule and product rule, the left-hand side involving the product \(e^y \, \cos x\) requires careful application of these rules.

The right side involves the function \(\sin(xy)\). Applying the chain rule, differentiate the outer function \(\sin\) to get \(\cos(xy)\), and multiply by the derivative of the inner function, \(xy\) to obtain \(y + x \frac{dy}{dx}\).

Combining these computations allows solving for \(\frac{dy}{dx}\). This process shows how to systematically approach more complex differentiations involving implicit functions.

An implicit function is one where the variable \(y\) is not isolated. It is written in relation to \(x\) within an equation, but not directly solved for \(y\). Such equations often require differentiation techniques that involve treating \(y\) as a function of \(x\) even if it’s not explicitly stated.

With implicit functions, you can't directly compute derivatives as you would with explicit functions, where one variable is entirely expressed in terms of the other.

• In implicit differentiation, both sides of an equation are differentiated with respect to \(x\).
• Then, solve for the derivative \(\frac{dy}{dx}\).

Implicit differentiation is essential when dealing with circles, ellipses, and other curves that aren't easily resolvable into \(y = f(x)\) form.

In our exercise, \(e^y \cos x = 1 + \sin(xy)\) illustrates this perfectly since it involves both \(x\) and \(y\) intertwined in various ways. This gives a glimpse into how we approach and solve problems involving implicit functions by finding the rate at which one variable changes with respect to another in such hidden forms.