Trigonometric Terms in Implicit Equations

Why Trigonometric Terms Need Care in Implicit Differentiation

When an equation relates \(x\) and \(y\), differentiating with respect to \(x\) means that \(y\) is treated as a function of \(x\). This matters especially when a trigonometric expression contains \(y\): its input changes as \(x\) changes, so the chain rule introduces a factor of \(\frac{dy}{dx}\).

The same idea applies when the input combines both variables. For example, the derivative of \(\sin(xy)\) involves the derivative of the entire input \(xy\), not just the derivative of the sine function. After differentiating every term, the result is usually a linear equation in \(\frac{dy}{dx}\). The goal is to collect those terms and solve for the slope.

The Chain Rule for Trigonometric Terms

For a differentiable input \(u=u(x)\), the familiar derivatives are \(\frac{d}{dx}\sin u=\cos u\cdot\frac{du}{dx}\), \(\frac{d}{dx}\cos u=-\sin u\cdot\frac{du}{dx}\), and \(\frac{d}{dx}\tan u=\sec^2u\cdot\frac{du}{dx}\). The tangent formula applies where \(\cos u\ne 0\), so that \(\tan u\) is defined.

In implicit differentiation, an input may depend on \(x\) both directly and through \(y\). If \(u=u(x,y)\) and \(y=y(x)\), the multivariable chain rule gives \(\frac{du}{dx}=u_x+u_y\frac{dy}{dx}\), where \(u_x\) and \(u_y\) are the partial derivatives of \(u\) with respect to its indicated variables. Consequently, the derivative of \(\sin(u(x,y))\) is \(\cos(u)(u_x+u_y\frac{dy}{dx})\). The corresponding cosine and tangent derivatives use \(-\sin(u)\) and \(\sec^2(u)\) in place of \(\cos(u)\).

These formulas are applications of the ordinary chain rule: first regard \(u(x,y(x))\) as a single function of \(x\), differentiate that input, and multiply by the derivative of the outer trigonometric function. They explain why an input involving \(y\) contributes a \(\frac{dy}{dx}\) factor.

A Reliable Differentiation Process

The procedure is the same whether the trigonometric input is \(y\), \(xy\), or a longer expression in both variables. Differentiate both sides with respect to \(x\), applying the chain rule to each trigonometric term. Then move all terms involving \(\frac{dy}{dx}\) to one side and all other terms to the other side. Finally, divide by the coefficient of \(\frac{dy}{dx}\), provided that coefficient is not zero.

Worked Examples

Why Collecting the Derivative Terms Works

After differentiation, each occurrence of \(\frac{dy}{dx}\) appears multiplied by a factor that may depend on \(x\) and \(y\). No product of two \(\frac{dy}{dx}\) terms is introduced by differentiating a relation once: the chain rule differentiates each input once, and the product rule differentiates one factor at a time. Thus, after simplifying, the equation has the form \(A(x,y)\frac{dy}{dx}=B(x,y)\), where \(A\) and \(B\) are expressions in the variables.

If \(A(x,y)\ne 0\), division gives \(\frac{dy}{dx}=\frac{B(x,y)}{A(x,y)}\). This is why collecting the derivative terms is not just a bookkeeping preference: it exposes the coefficient that must be nonzero to determine the slope in this way. If that coefficient vanishes, the rearranged equation cannot be divided by it; one must not use the resulting quotient at that point.

Common Mistake: Forgetting the Derivative of the Input

A frequent error is to differentiate \(\cos y\) as though its input were \(x\). The incorrect calculation is

This omits the derivative of \(y\) with respect to \(x\). Since \(y\) depends on \(x\), the chain rule gives the correct calculation:

The same check is useful for an input involving both variables. For example, \(\frac{d}{dx}\tan(x+y)\) is not just \(\sec^2(x+y)\); the derivative of the input must also be included:

Whenever differentiating a trigonometric term, identify its entire input first. Then differentiate that input before moving on to collect derivative terms.